Why the two expressions for number $e$ converge to the same number? We 'know' that
$$\lim_{n\to\infty}\left(1+\frac 1n\right)^n=\sum_{n=0}^\infty\frac 1{n!}$$
To prove this, I have started with the inequality
$$\left(1+\frac1n\right)^n=\sum_{k=0}^n\frac1{k!}\prod_{j=0}^{k-1}\frac{n-j}n\le\sum_{k=0}^n\frac1{k!}$$
Then I tried to find a strictly increasing sequence of natural numbers $a_n$ such that the following inequality holds for every $n$:
$$\left(1+\frac1{a_n}\right)^{a_n}\ge\sum_{k=0}^n\frac 1{k!}$$
I have tried with $a_n=n+1$ and $a_n=2n$, but I couldn't prove it. In fact, I tried some terms with a calculator and it seems to be false for these particular sequences of natural numbers. Is this the way to prove the equality? If it is not, how?
EDIT: I should have said this before, but exp and log functions are not already defined. So the answers to the 'duplicate' are useless to me.
 A: It is easy to show that both $\lim_{n\to\infty}\left(1+\frac1n\right)^n$ and $\sum_{k=0}^\infty \frac1{k!}$ exist.  We shall use this in that which follows.

First, you were correct to begin by using the binomial theorem to write
$$\begin{align}
\left(1+\frac1n\right)^n&=\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}\\\\
&=\sum_{k=0}^n \frac1{k!}\prod_{j=0}^{k-1} \left(1-\frac{j}{n}\right)\\\\
&\le \sum_{k=0}^n \frac1{k!}\tag1
\end{align}$$
Hence, we see that from $(1)$ that
$$\lim_{n\to\infty }\left(1+\frac1n\right)^n\le \sum_{k=0}^\infty \frac1{k!}\tag2$$

We also have for $2\le m\le n$
$$\begin{align}
\left(1+\frac1n\right)^n&=\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}\\\\
&\ge\sum_{k=0}^m \frac1{k!}\prod_{j=0}^{k-1} \left(1-\frac{j}{n}\right)\tag3
\end{align}$$
Letting $n\to\infty$ in $(3)$ reveals
$$\lim_{n\to\infty}\left(1+\frac1n\right)^n\ge \sum_{k=0}^m \frac1{k!}\tag4$$
whereupon letting $m\to \infty$ in $(4)$ yields
$$\lim_{n\to \infty}\left(1+\frac1n\right)^n\ge \sum_{k=0}^\infty \frac1{k!}\tag5$$

Putting together $(2)$ and $(5)$ provides proof of equality.  And we are done!
