Prove that $(\frac{n+1}{4})^{n+1} < n!$ by induction

Question

Prove that $(\frac{n+1}{4})^{n+1} < n!$ for integer $n\geq 1$ by induction

Answer 1

Base case is trivial.

By induction hypotheses we have $({{n+1}\over4})^{n+1} < n!$

Note that $(1+{1\over n})^n < e$ for all $n$ by definition of $e$.

So $({{n+2}\over4})^{n+2} = ({{n+2}\over4})({{n+2}\over{n+1}})^{n+1}({{n+1}\over4})^{n+1} < ({{n+2}\over4})\cdot e\cdot n! < (n+1)!$