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.
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!