Calculating the limit $\lim((n!)^{1/n})$

By rearranging terms, we can see that $$(n!)^2=[1\cdot n][2\cdot (n-1)][3\cdot (n-2)] \cdots [(n-1)\cdot 2][n\cdot 1].$$ Each of the $n$ products $(k+1)\cdot (n-k)$, for $0\le k<n$, is $\ge n$. Thus $$(n!)^2 \ge n^{n} \quad\text{and therefore}\quad (n!)^{1/n}\ge \sqrt{n}.$$

$$ \begin{aligned} \lim_{n\to\infty} (n!)^{1/n} &=\lim_{n\to\infty} \exp(\tfrac{1}{n} \ln n!)\\ &= \lim_{n\to\infty} \exp[\tfrac{1}{n} (\ln 1+\ln 2+\cdots + \ln n)]\\ &\ge \lim_{n\to\infty}\exp \left[ \frac{1}{n} \int_1 ^n \ln x dx\right]\\ &=\lim_{n\to\infty} \exp \frac{n\ln n -n+1}{n} \end{aligned} $$

and last side of above inequality diverges.

Taking $\log$ and using Stolz-Cesaro: $$ \log\lim_{n\to\infty}n!^{1/n}= \lim_{n\to\infty}\log n!^{1/n}= \lim_{n\to\infty}{\log 1+\cdots+\log n\over n}= \lim_{n\to\infty}{\log(n+1)\over(n+1)-n}= \lim_{n\to\infty}\log(n+1)=\infty,$$ so $\lim n!^{1/n}=\infty$.