Simplest or nicest proof that $1+x \le e^x$

Solution 1:

Another way (not sure if its "simple" though!): $y = x+1$ is the tangent line to $y = e^x$ when $x= 0$. Since $e^x$ is convex, it always remains above its tangent lines.

Solution 2:

$$ e^x = \lim_{n\to\infty}\left(1+\frac xn\right)^n\ge1+x $$

by Bernoulli's inequality.

Solution 3:

The shortest proof I could think of: $$1 + x \leq 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = e^x.$$

However, it is not completely obvious for negative $x$.

Using derivatives:

Take $f(x) = e^x - 1 - x$. Then $f'(x) = e^x - 1$ with $f'(x) = 0$ if and only if $x = 0$. But this is a minimum (global in this case) since $f''(0) = 1 > 0$ (the second derivative test). So $f(x) \geq 0$ for all real $x$, and the result follows.

Another fairly simple proof (but it uses Newton's generalization of the Binomial Theorem which is often covered in precalculus):

We proceed by contradiction. Suppose the inequality does not hold, i.e., $e^x < 1 + x$ for some $x$. Then $e^{kx} < (1 + x)^k$. Now set $x = 1/k$ so that \begin{align*} e &< \left( 1 + \frac{1}{k} \right)^k\\ &= 1 + \frac{k}{1}\left( \frac{1}{k} \right)^1 + \frac{k(k - 1)}{1 \cdot 2}\left( \frac{1}{k} \right)^2 + \frac{k(k - 1)(k - 2)}{1 \cdot 2 \cdot 3}\left( \frac{1}{k} \right)^3 + \cdots\\ &< 1 + \frac{k}{1}\left( \frac{1}{k} \right)^1 + \frac{k^2}{1 \cdot 2}\left( \frac{1}{k} \right)^2 + \frac{k^3}{1 \cdot 2 \cdot 3}\left( \frac{1}{k} \right)^3 + \cdots\\ &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots\\ &= e, \end{align*} which is absurd. Therefore $1 + x \leq e^x$ for all real $x$.

By the way, this is where $$e = \lim_{k \to \infty}\left( 1 + \frac{1}{k} \right)^k$$ comes from because $$\lim_{k \to \infty}\frac{k(k - 1)}{k^2} = \lim_{k \to \infty}\frac{k(k - 1)(k - 2)}{k^3} = \cdots = \lim_{k \to \infty}\frac{k(k - 1)(k - 2) \cdots (k - n)}{k^{n + 1}} = \cdots = 1.$$