calculus - Why is$(1+frac{1}{n})^n=e$ when n goes to infinity?

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

I think it involves $\sum\limits_{n=0}^\infty\frac1{k!}=e$ but not sure how to get from one to the other.

Answer

Have you tried expanding by the binomial theorem? ;-)
$$(1/n+1)^n=\sum_{k=0}^n\binom{n}k\frac1{n^k}=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac1{n^k}$$ then as $n\to\infty$ we find $n!/(n^k (n-k)!)\to1$ hence we have: $$\lim_{n\to\infty}\sum_{k=0}^n\frac{n!}{k!(n-k)!\cdot n^k}=\sum_{k=0}^\infty\frac1{k!}\equiv e$$