Is this inequality true? $ (nk)! \ge n^k \cdot k! $

Question

Is this inequality true? $$ (nk)! \ge n^k \cdot k! $$

I fixed a $k$ and tried a few values of $n$, and it seems to be true. But when trying to prove it by induction on $n$, I am not getting anywhere.

Bases case $n=1$ is true with equality. Assume it is true for some $n$, i.e. $ (nk)! \ge n^k \cdot k! $ is true. Now we need to show $ ((n+1)k)! \ge (n+1)^k \cdot k!$

$$ \left( \left( n+1 \right)k \right)!=\left( nk+k \right)!=\left( nk \right)!k!\left( \begin{matrix} nk+k \\ k \\ \end{matrix} \right)\ge {{n}^{k}}{{\left( k! \right)}^{2}}\left( \begin{matrix} nk+k \\ k \\ \end{matrix} \right) $$

But this gets me nowhere since I need a $n+1$ to appear after the inequality. Suggestions? I guess there must be an easier non-induction proof.

Answer 1

$$ (nk)!=1\cdot 2\cdot\ldots\cdot nk= \ldots\cdot n\cdot \ldots\cdot2n\cdot \ldots\cdot 3n\cdot \ldots\cdot kn\ge k!n^k $$

Answer 2

The symmetric group $S_{nk}$ has $$\frac{(nk)!}{n^k k!}$$ elements consisting of $k$ cycles of length $n$. So not only $n^kk!\le(nk)!$, but $n^kk!\mid(nk)!$.