How to prove $\lim\limits_{n \to \infty} (1+\frac1n)^n = e$?

Actually, the way things work out in mathematics usually is that we only prove that $x_n = (1+1/n)^n$ is a convergent sequence, and we define its limit to be $e$. Some people use other definitions for $e$ and show that it is equivalent to this definition, but there are many ways to do this that are logically equivalent.

(Just for the record, you don't "prove" limits, you compute them.)

One way to show that this sequence is convergent is to show that it is increasing and bounded above. This is a bit technical but works out quite well. To prove that $x_n$ is increasing, we show that $x_{n+1} \ge x_n$, which is equivalent to $x_{n+1}/x_n \ge 1$ : \begin{align} \frac{x_{n+1}}{x_n} & = \frac{ \left( 1 + \frac 1{n+1} \right)^{n+1} }{\left( 1 + \frac 1n \right)^n } = \left( 1 + \frac 1{n+1} \right) \left( \frac{ 1 + \frac 1{n+1} }{1 + \frac 1n} \right)^n = \left( \frac {n+2}{n+1} \right) \left( \frac{(n+2)n}{(n+1)^2} \right)^n \\ & = \left( \frac{n+2}{n+1} \right) \left( 1 + \frac{(n+2)n - (n+1)^2 }{(n+1)^2} \right)^n \\ & = \left( \frac{n+2}{n+1} \right) \left( 1 - \frac{1}{(n+1)^2} \right)^n \\ & \ge \left( \frac{n+2}{n+1} \right) \left( 1 - \frac n{(n+1)^2} \right) \\ & \ge \left( \frac{n+2}{n+1} \right) \left( \frac{n^2 + n + 1}{n^2 + 2n + 1} \right) = \frac{n^3 + 3n^2 + 3n + 2}{n^3 + 3n^2 + 3n + 1} \ge 1. \end{align}

The first inequality holds by Bernoulli's inequality ( I'm referring to $(1+x)^n \ge 1+nx$ for $x \ge -1$).

To prove that $x_n$ is bounded above, we use the binomial theorem :

\begin{align} \left(1+\frac 1n\right)^n & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \frac 1{n^k} = \sum_{k=0}^n \frac 1{k!} \frac{n!}{n^k(n-k)!} \\ & \le 1 + \sum_{k=1}^n \frac 1{k!} \le 1 + \sum_{k=1}^n \frac 1{2^{k-1}} = 1 + \frac{1- \frac 1{2^n}}{ 1- \frac 12} < 1 + 2 = 3. \\ \end{align} Note that I used the fact that $k! \ge 2^{k-1}$, which implies $\frac 1{2^{k-1}} \ge \frac 1{k!}$. (To see this, just write $k! = k \times \dots \times 2 \times 1 \ge 2 \times \dots \times 2 \times 1 = 2^{k-1}$.)

Another way would be to show that this sequence converges to $\sum_{k=0}^{\infty} \frac 1{k!}$ (which is another definition for $e$) by expanding $(1+1/n)^n$ with the binomial theorem.

Hope that helps,

The fact that it converges to a limit is already explained in other answers. Then to get an idea as to why this limit must match the definition of $e$:

We know that $\exp(x) = e^x$ is a function that starts with $\exp(0) = 1$, and grows with a rate that is equal to it's current value (ie, $\frac{d}{dx}e^x = e^x$). So a good way to find this, is to approximate it with a finite number of segments, where the rate at each segment is constant, and equal to the value at the beginning of the segment:

$\exp_n(0) = 1$

$\exp_n(t) = \exp_n(t - \frac{1}{n}) + \frac{1}{n}\exp_n(t - \frac{1}{n})$

$\exp_n(t) = (1 + \frac{1}{n})\exp_n(t - \frac{1}{n})$

Clearly, taking smaller and smaller segments will give a better approximation of the actual smooth function, with a constantly changing rate, and in the limit, it will match the definition of $\exp$. Unrolling the recursion for $\exp_n(1)$ clearly gives

$\exp_n(1) = (1 + \frac{1}{n})^n \exp_n(0) = (1 + \frac{1}{n})^n$

so

$\exp(1) = \lim_{n \rightarrow \infty} \exp_n(1) = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$

This is enough to prove that if the limit exists, then it matches the definition of $e$.

Just apply the binomial theorem then move things around a bit: $$\left(1+\tfrac{1}{n}\right)^n = \sum_{k=0}^n \frac{{n \choose k}}{n^k} = \sum_{k=0}^n \frac{1}{k!} \frac{n!/(n-k)!}{n^k}.$$

Now if you could just get rid of that $\frac{n!/(n-k)!}{n^k}$ term... (using the idea of $n \to \infty$ and $k$ is usually small in comparison)

edit Jonas makes a good point, I was implicity assuming the definition $e = \sum_{k = 0}^\infty \frac{1}{k!}$