Why did Euler use e to represent complex numbers?

From Euler we've learned that $z=re^{i\theta}$.

And it's easy to see that $|z|^2=r^2$, since $re^{i\theta}\times re^{-i\theta}=r^2$.

Why must we use e to represent these numbers correctly? It seems that I could arbitrarily choose a different exponent $z=r\pi^{i\theta}$ and get the same size for $z$ as I did before: $|z|^2=r\pi^{i\theta}\times r\pi^{-i\theta}=r^2$

What did I miss?

If we wish to express $\pi^{i\theta}$ as a series then we have:

$$\pi^{i\theta} = e^{i\ln(\pi)\theta} = \sum_{n=0}^\infty i^n \frac{(\ln(\pi)\theta)^n}{n!} = \cos(\ln(\pi)\theta)+i\sin(\ln(\pi)\theta).$$

Calculating precisely $\ln(N)$ for $N \in \mathbb{N}$ can be difficult, not to mention $\ln(\pi)$. This would add more complications than it would be worth. Moreover, $\pi^{i\theta}$ has period $2\pi/\ln(\pi)$, which is not compatible with polar coordinates.

On the other hand, since we can write $$e^{i\theta} = \cos(\theta) + i \sin(\theta),$$ we can express $e^{i\theta}$ by calculating the already well known trigonometric functions.

I would like to add that the use of $e^{i\theta}$ is because of the nice representation found by Euler. If you were to approach the polar representation for the first time, you would approach it more like this:

Let $z=x+iy$ be a complex number, which we can visualize as a vector in $\mathbb{R}^2$, $z=(x,y)$. The magnitude of $z$ is $\|z\|= \sqrt{x^2+y^2}$. We can write the real part as $x=\|z\| \cos(\theta)$ where $\theta$ is the angle formed between the real axis and the vector at the origin. Similarly $y=\|z\| \sin(\theta)$. Thus $$z= \|z\|\cos(\theta)+i \|z\|\sin(\theta) = \|z\|(\cos(\theta)+i\sin(\theta)).$$

Until now, our reasoning was completely geometric. Independently we can work out the expression, due to Euler, $e^{i\theta} = \cos(\theta)+i\sin(\theta)$. This now naturally leads to $$z=\|z\|e^{i\theta}.$$ If it turned out that $\pi^{i\theta} = \cos(\theta)+i\sin(\theta)$ then we would use that instead. However, we know that this is not the case.

I would also like to point out that there is an intuitive reason to think that $e^{i\theta}$ should be of the form $\cos(\theta)+i\sin(\theta)$.

Notice that if we write $f(\theta) = e^{i\theta} = u(\theta)+iv(\theta)$, then $$f''(\theta) = i^2 f(\theta) = - f(\theta).$$

Hence $$u''(\theta) = -u(\theta) \text{ and } v''(\theta) = - v(\theta).$$

Thus from differential equations, we can express $u$ and $v$ as a linear combination of $\sin(\theta)$ and $\cos(\theta)$.

This motivates the investigation into the series of the exponential function. From this perspective, it is not surprising to discover $\cos(\theta)$ and $\sin(\theta)$ inside the series for $e^{i\theta}$.

One final edit: If we let $A$ and $B$ be complex numbers, then my previous statement can be expressed as: $$e^{i\theta} = A\cos(\theta)+B\sin(\theta)$$

Setting $\theta=0$ we see that $e^{0}=1=A\cdot 1 = A$. And $\theta = \pi/2$ yields $e^{i\pi/2} = B$.

Therefore, $$e^{i\theta} = \cos(\theta) + e^{i\pi/2} \sin(\theta).$$ What is left is to determine $e^{i\pi/2}$. Since $e^{i\theta}$ is $2\pi$ periodic, $e^{0}=e^{i2\pi}$. Thus we can see that $(e^{i\pi/2})^4 -1 = 0$, which means $e^{i\pi/2}$ satisfies the polynomial $x^4-1=0$. Thus $e^{i\pi/2} = \pm 1 \text{ or } \pm i$.

Taking the derivative of both sides of $e^{i\theta} = \cos(\theta) + e^{i\pi/2} \sin(\theta)$ we find: $$ie^{i\theta} = -\sin(\theta) + e^{i\pi/2} \cos(\theta)$$ and therefore by setting $\theta = 0$ we have: $$i = e^{i\pi/2}.$$ Thus we conclude $$e^{i\theta} = \cos(\theta)+i\sin(\theta).$$ All without Taylor series.

There is a nice formula for $e^x$, and only $e^x$:
$$e^x=1+x+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{24}+\frac{x^5}{120}+...$$
If you calculate $e^{0.1},e^{0.01}$, you can see the first couple of terms are correct.
So $$e^{ix}=1+ix+\frac{(ix)^2}2+\frac{(ix)^3}6+...\\=(1-\frac{x^2}2+\frac{x^2}{24}+....)+i(x-\frac{x^3}6+\frac{x^5}{120}-...)$$
Now, in radians, $$\cos x = 1-\frac{x^2}2+\frac{x^4}{24}...\\ \sin x = x-\frac{x^3}6+\frac{x^5}{120}...$$ You can check those for small values of $x$ as well. So the series for $e^{ix}$ and the series for $\sin$ and $\cos$ match (at least for small $x$).
If you do the calculus, you can find they match all the way down.

I think the answer is a matter of aesthetics. First, you have to make some assumptions, the most important of which is the following:

$$ e^{i\theta} = \cos(\theta) + i\sin(\theta) $$

You can "justify" this through a Maclaurin series expansion of $e^x$, $\sin(x)$ , and $\cos(x)$. But that really begs the question--you really need to prove that $\frac{d}{dx}e^x = e^x$ for the specific base of $e$ and $e$ alone or, conversely, that $\frac{d}{dx}\log_e(x) = \frac{1}{x}$ (the derivative of $\ln(x)$) for the specific logarithm with base $e$--you need to prove one on its own because using one to prove the other is circular! In my opinion, this proof--that $\frac{d}{dx}e^x = e^x$ or $\frac{d}{dx}\ln(x) = \frac{1}{x}$--is the only real answer to the question because that proof and that proof alone actually explains what is so special about the value of $e$.

I'm going to assume that we now agree that $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. A computer doesn't care if you prefer the base $\pi$ (i.e. $\pi^{i\theta} = \cos(\ln(\pi)\theta) + i\sin(\ln(\pi)\theta)$. However if we choose such a base for our complex numbers, then we no longer get the nice trigonometric properties.

Specifically, let's say that we have $z = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}$. If we use the original formulation of $z = e^{i\theta}$ then we immediately recognize (if we are good at trigonometry) that $\theta = 45^\circ = \frac{\pi}{4}$. On the other hand, what is $\theta$ if we prefer $\pi^{i\theta}$? It's not clear what the value is--in fact we basically have to re-engineer the value--because we know the following:

\begin{align} z = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} =&\ \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) \\ =&\ \cos\left(\ln(\pi)x\right) + i\sin\left(\ln(\pi)x\right) \end{align}

Leading to:

$$ x = \frac{\pi}{4\ln(\pi)} \approx 0.68609911657 $$

This value has no geometric meaning--but it is no less correct mathematically. The approximate value of $0.68609911657$ rad is $\approx 0.21\pi$. Now, if I know that this is a bad base, then I can take that value and multiply by $\ln(\pi)$ to get: $0.68609911657 * \ln(\pi) \approx 0.78539816339$ which is $\approx 0.24999999999 \pi$ (which is clearly $\frac{\pi}{4}$).

With the initial representation I can quickly find where to place my coordinate on a complex plane where $x$ is the real values and $y$ are the imaginary values using polar coordinates. I can "quickly" do the same using $\pi^{i\theta}$ but it requires some intermediate steps to convert to the "nice" representation (which is $e^{i\theta}$). If, to use polar coordinates, I need to "convert" $\pi^{i\theta}$ then it makes more sense to choose the "natural" coordinates which are achieved using rather $e^{i\theta}$--where no conversion is necessary.

And the reason I say it's about aesthetics is because it's not like $e^{i\theta}$ makes all calculations easier--it doesn't. Try to find $r$ and $\theta$ from $z = 1 + 2i$:

\begin{align} 1 + 2i = r \cos(\ln(\pi)\theta) + ri\sin(\ln(\pi)\theta) \\ 1 + 2^2 = r^2 \rightarrow r = \sqrt{5} \\ \tan(\ln(\pi)\theta) = 2 \rightarrow \ln(\pi)\theta \approx 1.10714872 + 2\pi n\\ \theta \approx 0.96717027631 + \frac{2\pi n}{\ln(\pi)} \\ 1 + 2i \approx \sqrt{5}\pi^{\left(0.96717027631 + \frac{2\pi n}{\ln(\pi)}\right)i} \end{align}

vs.

\begin{align} 1 + 2i = r \cos(\theta) + ri\sin(\theta) \\ 1 + 2^2 = r^2 \rightarrow r = \sqrt{5} \\ \tan(\theta) = 2 \rightarrow \theta \approx 1.10714872 + 2\pi n\\ 1 + 2i \approx \sqrt{5}e^{\left(1.10714872 + 2\pi n\right)i} \end{align}

The only difference in the above calculations is the divide by $\ln(\pi)$--and that is why we prefer $e^{i\theta}$ over any other base--because all of the others require this "unnecessary" step--it's more aesthetically pleasing but no more mathematically correct.

And before you say, well at least using $e^{i\theta}$ you immediately know that the trigonometric angle is approximately $1.10714872$ radians, I would argue that I am actually not familiar with what that angle represents--in fact I need to convert to degrees to show that $1.10714872 \approx 63.4349489493^\circ$ before I really "know" where that value lies. And I can get that value just the same using the $\pi^{i\theta}$ it just requires an additional multiply by $\ln(\pi)$: $\theta \approx 0.96717027631 * \ln(\pi) * \frac{180^\circ}{\pi} \approx 63.4349489492^\circ$.

Once we know $e^{i\theta} = \cos\theta + i \sin\theta$, the expansion of a complex number into $re^{i\theta}$ then becomes natural. What is the idea behind the polar form? It is to express a complex $z$ in terms of a magnitude and direction. What do we mean by direction? Well, in general a "direction" is just a unit vector. We use this term because multiplying by "just a direction" shouldn't change any magnitudes. This tells us that unit vector is a good definition of "direction". How can we write unit vectors then? We just said how! The unit circle is parametrized by $\cos\theta + i \sin\theta$. We now have no choice. We must pick $e$ and not $\pi$ as you question. The math has decided for us that $\cos\theta + i\sin\theta = e^{i\theta}$, not $\pi^{i\theta}$. Once we have settled on how we are going to express directions, now what does our polar form dictate the corresponding magnitude should be? The polar magnitude of $z$ has to be $|z|$, how convenient!