Prove nth root of n! is less than n+1 th root of ((n+1) !): $\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$? 
Prove the inequality $$\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$$ 


As $n!$ is only defined if $n\in \mathbb{N}\cup \{ 0\}$, use logarithms. You can make: 
$$\frac{1}{n} \log(n!)< \frac{1}{n+1}\log((n+1)!)$$
$$\Leftrightarrow (n+1)\log(n!)< n\log((n+1)!)$$
$$\Leftrightarrow (n+1)\log(n!)< n\log(n!)+ n\log(n+1)$$
$$\Leftrightarrow n\log(n!)+\log(n!)< n\log(n!)+ n\log(n+1)$$
$$\log(n!)< n\log(n+1)$$
$$\Leftrightarrow \log(n!)<log((n+1)^n)$$
And that is true because $n!<(n+1)^n$ noting that you have $n$ times $n+1$, which is greater than the product of the numbers from 1 to $n$. 
