I have to find the following limit but I don't know where to begin with.
$$\lim_{n\to\infty}\frac{n^k\;(k+1)!\;(n-(k+1))!}{n!}$$ for any fixed $k$.
Any hints or suggestions would help me a lot. I think sandwich rule will be useful, but I cannot think of a suitable expression.
If I've counted all my $n$'s correctly here, we will have:
$$\begin{array}{rcl}\lim_{n\to\infty}\frac{(n^k)(k+1)!(n-(k+1))!}{(n)!}&=&(k+1)!\lim_{n\to\infty}\frac{1}{n}\frac{n^{k+1}(n-(k+1))!}{n!}\\&=&(k+1)!\lim_{n\to\infty}\frac{1}{n}\times\frac{n}{n}\frac{n}{n-1}\cdots\frac{n}{n-k}\\&=&0\end{array}$$
because all terms $\frac{n}{n-i}\to 1$ but $\frac{1}{n}\to 0$ as $n\to\infty$.
We have that
$$\frac{(n^k)(k+1)!(n-(k+1))!}{(n)!}=\frac{n^k}{\binom{n}{k+1}} \to 0$$
indeed by this result $\binom{n}{k+1}=\frac{n^{k+1}}{(k+1)!}+O(n^{k})$ then
$$\frac{n^k}{\binom{n}{k+1}}=\frac{n^k}{\frac{n^{k+1}}{(k+1)!}+O(n^{k})}=\frac{1}{\frac{n}{(k+1)!}+O(1)}\to 0$$
