Proof for 1/k! using n choose k as n approaches infinity and its relation to the gamma function

Question

Prove that $\lim_{ n \to \infty }\binom{n}{k}(1/n)^k =\frac{1}{k!}$

How is this related to the gamma function?

Answer 1

$$\frac{1}{n^k}\binom{n}{k} = \frac{1}{k!}\frac{n!}{n^k(n-k)!} = \frac{1}{k!}\prod_{j=0}^{k-1}\frac{n-j}{n} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)$$ and every term of the last product tends to one as $n\to +\infty$. Moreover, $$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}.$$

Answer 2

$\displaystyle lim_{n \rightarrow \infty}{n \choose k}\frac{1}{n^k} = \frac{1}{k!} lim_{n\rightarrow \infty}\frac{\frac{n!}{(n-k)!}}{n^k} = \frac{1}{k!}$

It turns out that this result can be used to derive a Taylor series, as I show in this post Why does a power series with binomial coefficients reduce to a Taylor series?

$f(t)= \displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^{n}{n \choose k} \left( \frac{t}{n}\right)^kf^{(k)}(0)= \sum_{k=0}^{\infty}\frac{f^{(k)}(0)t^k}{k!}\lim_{n\rightarrow \infty}\frac{\frac{n!}{(n-k)!}}{n^k}=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)t^k}{k!}$