<s>[INST] How to prove $\lim\limits_{n \to \infty} (1+\frac1n)^n = e$? 
How to prove the following limit?
  $$\lim_{n \to \infty} (1+1/n)^n = e$$ 
  I can only observe that the limit should be a very large number!

Thanks.
 [/INST] 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,
</s>