discrete mathematics - Use induction to prove that $n! leq n^{n-1}$

Use induction to prove that $n! \leq n^{n-1}$ for all integers $n\geq 1$. I'm having a hard time with induction and my professor said this is a good future test like question if someone can post a solution and explain it would help me out a lot. Thank you.

Answer

Call $n!\le n^{n-1}$ the statement $T(n)$.

$T(1)$ holds, because $1!\le 1^{1-1}=1^0=1$.

Now assume $T(n)$, i.e. $n!\le n^{n-1}$.
Then $$(n+1)!=n!(n+1)\le n^{n-1}(n+1)\le (n+1)^{n-1}(n+1)=(n+1)^n$$

So $T(n+1)$ holds too.

We've proved $T(1)$ and that $\forall n\in\mathbb N(T(n)\implies T(n+1))$. $\ \ \ \square$