Originally Posted by

**Laurent** I am not sure this is fully rigorous: for instance, don't you need to prove first that the limit exists? (and a few steps would need justifications, like $\displaystyle \lim_{n\to\infty}(n!)^{\frac{-1}{n(n+1)}}=\lim_{n\to\infty}\frac{1}{(n!)^{\frac{ 1}{n^2}}}$ or the previous one)

I would rather say: readily, $\displaystyle 1\leq n!=n(n-1)\cdots 1\leq n^n$, hence $\displaystyle 1\leq (n!)^{\frac{1}{n^2}}=\exp(\frac{\log (n!)}{n^2})\leq \exp(\frac{\log n}{n})$. Since $\displaystyle \frac{\log n}{n}\to 0$ (and $\displaystyle \exp$ is continuous at 0), the limit indeed exists and equals 1.