Prove $\sqrt n\leq \sqrt[n]{n!} \leq \frac{n+1}{2}$.
I thought I would use induction to prove this, only I don't know how to use the assumption in the inductive step.
Base: $n=1$ $\Rightarrow \sqrt 1\leq \sqrt[1]{1!}\leq \frac{2}{2}$. So for $n=1$ this inequality holds.
Assumption: $\sqrt k\leq \sqrt[k]{k!}\leq \frac{k+1}{2}$
Inductive step: $\sqrt{k+1}\leq \sqrt[k+1]{(k+1)!}\leq \frac{k+1+1}{2}$
Is it correct so far and how do I continue from here?
