real analysis - Prove $2^ncdot n! ≤ (n+1)^n$ by induction.

An induction I'm struggling with.

Prove $2^n\cdot n! ≤ (n+1)^n$ by induction.

An idea was to show that $2^n\cdot n! ≤ 1+n^2$ since $1+n^2 ≤ (n+1)^n$ using Bernoulli. However the inequality is just wrong so that approach doesn't work. I had the intuition that $2^n ≤ n!$ but I don't think that yields anything for this problem.

I would really like to get a hint or two. Of course you can post your answer, this is obviously what this platform is for, but I won't read them until I solved the problem myself. It's an induction, can't be that difficult right?

Answer

Hint: The induction step goes as follows:
$$2^{n+1}(n+1)!=2^nn!2(n+1)\le(n+1)^n2(n+1)=2(n+1)^{n+1}$$
Thus you are left to prove that $2(n+1)^{n+1}\le(n+2)^{n+1}$, which is pretty easy.