Induction problem, how to prove the final step

Question

$\displaystyle\sqrt[n]{n!}\leq\frac{n+1}{2}$ for all $n\in\mathbb{Z}^+$

Answer 1

If $m!\le\left(\dfrac{m+1}2\right)^m$

$(m+1)!=(m+1)\cdot m!\le(m+1)\cdot\left(\dfrac{m+1}2\right)^m$

It is sufficient to show $$(m+1)\cdot\left(\dfrac{m+1}2\right)^m\le\left(\dfrac{m+2}2\right)^{m+1}\iff\left(1+\dfrac1{m+1}\right)^{m+1}\ge2$$

Answer 2

$$\left(1+\dfrac1{m+1}\right)^{m+1}\ge2$$ to show easily take $n=m+1$ so $$\left(1+\dfrac1{n}\right)^{n}\ge2\\ \left(\begin{array}{c}n\\ 0\end{array}\right)1^n\dfrac 1n^0+ \left(\begin{array}{c}n\\ 1\end{array}\right)1^{n-1}\dfrac 1n^1+ \left(\begin{array}{c}n\\ 2\end{array}\right)1^{n-2}\dfrac 1n^2+...=\\ 1+n.1^{n-1}.\dfrac 1n+\dfrac{n(n-1)}{2}.1^{n-2}.\dfrac{1}{n^2}+...=\\ 1+1+\dfrac{n-1}{2n}+...\geq2$$