Prove $\sqrt[n]{n!}\le{{n+1}\over2}$, $n \in N^+$, using induction.
This is how far I got, but then I got stuck:
$$\sqrt[n+1]{(n+1)!}\le{{n+2}\over2}$$ $$((n+1)!)^{1\over{n+1}}\le{n\over2}+1$$ $$((n+1)\times n!)^{1\over{n+1}}\le{n\over2}+1$$ $$(n+1)^{1\over{n+1}}\times n!^{1\over{n+1}}\le{n\over2}+1$$ Can i somehow substitute $(n+1)^{1\over{n+1}}$ with ${{{n+2}\over2}}$ using the original proposition $\sqrt[n]{n!}\le{{n+1}\over2}$?
How would I continue?