Fdtxjr

I'm kind of confused, because I'm not sure how this expression
$$
sum_k=0^n fracn!k!(n-k)!n^k
$$
simplifies to $$
sum_k=0^n frac1k!
$$

Could anybody help me out?

As Clement C. already noted, the two expressions are not equivalent. Here, we show that, in fact, $left(1+frac1nright)^nle sum_k=0^n frac1k!$ for all $nge 1$. Moreover, we show that as $nto infty$, both terms of interest approach the same value. To that end, we proceed.

First, note that using the binomial theorem, we can write

$$beginalign
left(1+frac1nright)^n&=sum_k=0^n binomnkfrac1n^k\\
&=sum_k=0^n frac1k!fracn!n^k(n-k)!\\
&=sum_k=0^n frac1k!fracn(n-1)(n-2)cdots (n-k+1)n^k\\
&=sum_k=0^n frac1k!left(frac nnright)left(fracn-1nright)left(fracn-2nright)cdots left(fracn-k+1nright)\\
&=sum_k=0^n frac1k!left(1-frac1nright)left(1-frac2nright)cdots left(1-frack-1nright)tag 1
endalign$$

Second, since all of the parenthetical terms in $(1)$ are less than $1$, we arrive at the first coveted relationship

$$bbox[5px,border:2px solid #C0A000]left(1+frac1nright)^n le sum_k=0^nfrac1k! tag 2$$

for all $nge 1$.

Next, note that since all of the terms in the binomial expansion are positive, we have for any integer $0le p<n$

$$left(1+frac1nright)^n ge sum_k=0^p binomnkfrac1n^k tag 3$$

Putting $(2)$ and $(3)$ together yields

$$sum_k=0^p binomnkfrac1n^k le left(1+frac1nright)^n le sum_k=0^nfrac1k! tag 4$$

Since $left(1+frac1nright)^n$ is monotonically increasing (SEE THIS ANSWER) and bounded above by the convergent series $sum_k=0^infty frac1k!$, 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

$$sum_k=0^p frac1k! le lim_nto inftyleft(1+frac1nright)^n le sum_k=0^inftyfrac1k! tag 5$$

for all $p$. Since the inequality on the left-hand side of $(5)$ is true fo all $p$, we can let $pto infty$ to reveal the coveted result

$$bbox[5px,border:2px solid #C0A000]lim_nto inftyleft(1+frac1nright)^n=sum_k=0^infty frac1k!$$

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 $ntoinfty$, namely $e$.

For the first, it is trivial based on the series definition of $e$:
$$
sum_k=0^n frac1k! xrightarrow[ntoinfty] sum_k=0^infty frac1k! = e
$$

For the second, observe that by the Binomial theorem
$$
sum_k=0^n fracn!k!(n-k)! frac1n^k
= sum_k=0^n binomnkfrac1n^k
= left(1+frac1nright)^n xrightarrow[ntoinfty] e
$$
where the last step is a standard limit (or, for some, this is the definition of $e$).