Factorial Limit So I'm trying to prove the following: $$\lim_{n\rightarrow \infty } \sum_{k=0}^{n} \frac{1}{k!}= \lim_{n\rightarrow \infty }\sum_{k=0}^{n} \binom{n}{k}\frac{1}{n^k}, n\geq k$$
I've taken the assumption that given the summation of two limits extending to infinity over the same variable $n$, I take that 
$$\frac{1}{k!} \approx \frac{n!}{(n-k)!k!n^{k}}$$ with some light expansion of the right-hand side and shifting of terms i have that 
$$\frac{1}{k!} \approx \frac{n!}{(n-k)!n^{k}}\cdot \frac{1}{k!}$$ 
Which implies that:
$$\lim_{n\rightarrow \infty} \frac{n!}{(n-k)!n^{k}}=1 \Rightarrow or \Rightarrow \lim_{n\rightarrow \infty} \frac{(n-1)(n-2)...(n-k+1)}{n^{k-1}}=1$$
This is where I got stuck..any help is much appreciated in advance. Thanks
 A: $\sum_\limits{k=0}^n {n\choose k} \frac {1}{n^k}$ is the expansion of $(1+\frac 1n)^n$
We might write it out like so:
$1 + n(\frac 1n) + \frac {n(n-1)}{2}(\frac 1{n^2})+\cdots\\
1 + 1 + \frac 12 (1-\frac 1n)+ \frac 1{3!}(1-\frac 1n)(1-\frac 2n)\cdots$ 
We can match term by term with $\sum_\limits{k=0}^n \frac {1}{k!}$
When $n$ is finite, every term in 
$\sum_\limits{k=0}^n {n\choose k} \frac {1}{n^k}$ is less  than its corresponding term in $\sum_\limits{k=0}^n  \frac {1}{k!}$
When $n$ is finite 
$\sum_\limits{k=0}^n {n\choose k} \frac {1}{n^k} < \sum_\limits{k=0}^n  \frac {1}{k!}$
Consider, $\sum_\limits{k=0}^m {n\choose k} \frac {1}{n^k}$ with $m<n$ i.e. the first $m$ terms of the expansion of $(1+\frac 1n)^n$ 
$1 + 1 + \frac 12 (1-\frac 1n)+ \frac 1{3!}(1-\frac 1n)(1-\frac 2n)+\cdots +\frac 1{m!}(1-\frac 1n)(1-\frac 2n)\cdot(1-\frac {m-1}n)$
letting $n$ get big
$\lim_\limits{n\to \infty} \sum_\limits{k=0}^m {n\choose k} \frac {1}{n^k} = \sum_\limits{k=0}^m \frac 1{m!}$
And then we get let $m$ get to be as big as we need it to be.
