A proof by induction on $n$, instead of $k$, does work. Suppose that the inequality holds for certain $n$. Then for $n + 1$ we want to prove the marked inequality below.
$\binom{n+1}{k} = \frac{n+1}{n-k+1} \binom{n}{k} \leq \frac{n+1}{n-k+1} \cdot \frac{n^n}{k^k (n - k)^{(n - k)}} \stackrel{?}{\leq} \frac{(n+1)^{n+1}}{k^k (n - k + 1)^{(n - k + 1)}}$
Rewriting this marked inequality gives
$\left(\frac{n-k+1}{n-k}\right)^{n-k} \stackrel{?}{\leq} \left(\frac{n+1}{n}\right)^{n}$
Writing $n - k = m < n$, we get
$\left(1 + \frac{1}{m}\right)^m \stackrel{?}{\leq} \left(1 + \frac{1}{n}\right)^n$
Since $f(x) = (1 + \frac{1}{x})^x$ is increasing in $x$ (with limit value $e$) the result then follows.