Proof that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$ In a Calculus textbook, I was faced with the following
Problem: If $n$ is a natural number with $n \ge 1$, proof that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$ for all $k \in \mathbb{N}$.
The book presented a different solution than the one, I came up with. So maybe mine is wrong. My question now is, if my solution is correct and - more importantly - if not, where I went wrong. Thanks in advance for any comments and advice.
Solution: I proof the proposition by induction on $k$. For $k=0$ we have
$$\binom{n}{0}\frac{1}{n^0} = 1 \le \frac{1}{0!},$$
a true statement. Now notice, that if $k \gt n$ we have $\binom{n}{k} = 0$, so the proposition is plainly true. It thus suffices to proof it for $0 \le k \le n$. In this case we have as inductive step
$$\begin{align} \binom{n}{k+1} \frac{1}{n^{k+1}} & = \binom{n}{k} \frac{1}{n^{k}} \cdot \frac{1}{n(k+1)(n-k+1)} \\ & \le \frac{1}{k!} \cdot \frac{1}{n(k+1)(n-k+1)} && \text{by inductive hypothesis} \\ & \le \frac{1}{k!}\frac{1}{k+1} = \frac{1}{(k+1)!} && \text{since $n \ge 1$ and $n-k+1 \ge 1$}\end{align}$$
as required. $\blacksquare$
As additional comment, I have used the following steps to manipulate the binomial coefficient:
$\binom{n}{k+1} = \frac{n!}{(k+1)!(n-k+1)!} = \frac{n!}{(k+1)k!(n-k+1)(n-k)!} = \frac{n!}{k!(n-k)!} \cdot \frac{1}{(k+1)(n-k+1)} = \binom{n}{k} \frac{1}{(k+1)(n-k+1)}$

There is a mistake in your derivation: $$\binom{n}{k+1} = \frac{n!}{(k+1)!(n-k\color{red}-1)!} \color{blue}{= \frac{n!}{k!(n-k)!} \cdot \frac{n-k}{k+1} = \binom{n}{k} \cdot \frac{n-k}{k+1}}.$$
So, the final step of your proof should be the following:
\begin{align} 
\binom{n}{k+1} \frac{1}{n^{k+1}} & = \binom{n}{k} \frac{1}{n^{k}} \cdot  \frac{n-k}{n(k+1)} \\ 
& \le \frac{1}{k!} \cdot \frac{n-k}{n(k+1)} && \text{by induction hypothesis} \\ 
& \le \frac{1}{(k+1)!}\frac{n-k}{n} \\
& < \frac{1}{(k+1)!} && \text{since $n-k < n$}
\end{align}
