How does the binomial expansion of $(1+(1/n))^n$ equal to the partial sum sequence of $(1/n!)$? I'm kind of confused, because I'm not sure how this expression
$$
\sum_{k=0}^n \frac{n!}{k!(n-k)!n^k}
$$
simplifies to $$
\sum_{k=0}^n \frac{1}{k!}
$$
Could anybody help me out?
 A: 
As Clement C. already noted, the two expressions are not equivalent.  Here, we show that, in fact, $\left(1+\frac1n\right)^n\le \sum_{k=0}^n \frac{1}{k!}$ for all $n\ge 1$.  Moreover, we show that as $n\to \infty$, both terms of interest approach the same value.  To that end, we proceed.

First, note that using the binomial theorem, we can write 
$$\begin{align}
\left(1+\frac1n\right)^n&=\sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}\\\\
&=\sum_{k=0}^n \frac{1}{k!}\frac{n!}{n^k(n-k)!}\\\\
&=\sum_{k=0}^n \frac{1}{k!}\frac{n(n-1)(n-2)\cdots (n-k+1)}{n^k}\\\\
&=\sum_{k=0}^n \frac{1}{k!}\left(\frac nn\right)\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots \left(\frac{n-k+1}{n}\right)\\\\
&=\sum_{k=0}^n \frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\tag 1
\end{align}$$
Second, since all of the parenthetical terms in $(1)$ are less than $1$, we arrive at the first coveted relationship
$$\bbox[5px,border:2px solid #C0A000]{\left(1+\frac1n\right)^n \le \sum_{k=0}^n\frac{1}{k!}} \tag 2$$
for all $n\ge 1$.
Next, note that since all of the terms in the binomial expansion are positive, we have for any integer $0\le p<n$
$$\left(1+\frac1n\right)^n \ge \sum_{k=0}^p \binom{n}{k}\frac{1}{n^k} \tag 3$$
Putting $(2)$ and $(3)$ together yields
$$\sum_{k=0}^p \binom{n}{k}\frac{1}{n^k} \le \left(1+\frac1n\right)^n \le \sum_{k=0}^n\frac{1}{k!} \tag 4$$
Since $\left(1+\frac1n\right)^n$ is monotonically increasing (SEE THIS ANSWER) and bounded above by the convergent series $\sum_{k=0}^\infty \frac{1}{k!}$, its limit exists.  Similarly, the sum on the left-hand side of $(4)$ is monotonically increasing in $n$ and bounded above.  Therefore, we may take the limit of $(4)$.  Proceeding, we find
$$\sum_{k=0}^p \frac{1}{k!} \le \lim_{n\to \infty}\left(1+\frac1n\right)^n \le \sum_{k=0}^\infty\frac{1}{k!} \tag 5$$
for all $p$.  Since the inequality on the left-hand side of $(5)$ is true fo all $p$, we can let $p\to \infty$ to reveal the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\left(1+\frac1n\right)^n=\sum_{k=0}^\infty \frac{1}{k!}}$$
A: tl;dr: the two are not equal, but have the same limit when $n$ goes to $\infty$.

First: You cannot "simplify the first to get the second." The two expressions are not equal.
Compute the first few terms to check that. I am lazy, so asked a software program to do it for me. The ratio of their first 10 terms (obtained with Mathematica):
 N[Table[(Sum[1/(k!), {k, 0, n}]/Sum[n!/(k! (n - k)! n^k), {k, 0, n}]), {n, 1, 10}] ]


{1., 1.11111, 1.125, 1.10933, 1.09177, 1.0779, 1.06745, 1.05943,
  1.05312, 1.04802}

A plot (ditto)
DiscretePlot[{Sum[1/(k!), {k, 0, n}], Sum[n!/(k! (n - k)! n^k), {k, 0, n}]}, {n, 1, 100}]


However, the two expressions do have the same limit when $n\to\infty$, namely $e$.
For the first, it is trivial based on the series definition of $e$:
$$
\sum_{k=0}^n \frac{1}{k!} \xrightarrow[n\to\infty]{} \sum_{k=0}^\infty \frac{1}{k!} = e
$$
For the second, observe that by the Binomial theorem
$$
\sum_{k=0}^n \frac{n!}{k!(n-k)!} \frac{1}{n^k}
= \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k}
= \left(1+\frac{1}{n}\right)^n \xrightarrow[n\to\infty]{} e
$$
where the last step is a standard limit (or, for some, this is the definition of $e$).
