I found a different proof of this fact avoiding Stirling.
Note $f(x) = (\frac{ex}{k})^k$ is a $C^2$ strictly convex function.
So $f(x) + f'(x)h < f(x+h)$ for $0 < h$
In particular, letting $h = 1$ we get
$$f'(x-1)+ f(x-1) < f(x)$$
$$(\frac{e(x-1)}{k})^{k-1}e + (\frac{e(x-1)}{k})^k< (\frac{ex}{k})^k$$
Noting that $(\frac{k}{k-1})^{k-1} < e$ since the ratio limits to $e$ from below. Substituting in the LHS for the second $e$, we get
$$(\frac{e(x-1)}{k})^{k-1}(\frac{k}{k-1})^{k-1} + (\frac{e(x-1)}{k})^k< (\frac{ex}{k})^k$$
$$(\frac{e(x-1)}{k-1})^{k-1} + (\frac{e(x-1)}{k})^k< (\frac{ex}{k})^k$$
Now the result follows by induction on $n+k$ for $k\lt n$, and Pascal's formula.
$\binom {n-1}{k} + \binom {n-1}{k-1} = \binom {n}{k}$
$ \binom {n-1}{k} \le (\frac{e(n-1)}{k})^k$ and $\binom {n-1}{k-1} \le (\frac{e(n-1)}{k-1})^{k-1}$ imply $\binom{n}{k} \le (\frac{en}{k})^k$
The base case $\binom{n}{n}$ and $\binom{n}{0}$ and $\binom{n}{1}$ are trivial so we can avoid the technicality where $k=1$ and $k=0$.