TITLE: What is the limit as k approaches infinity of $(k!)^{\frac{1}{k}}$
QUESTION [7 upvotes]: What is the value of
$$\lim_{k\to\infty}(k!)^{\frac{1}{k}}?$$
One of my students concluded the limit was infinity – which I tend to agree with, but was unable to show that was the limit.  We knew  $k!$ was tough to beat, but $k^k$ does – so this situation was unclear.

REPLY [3 votes]: We have $k! < k^k$. Further, we have
$$e^k = \sum_{l=0}^{\infty} \dfrac{k^l}{l!} \implies e^k > \dfrac{k^k}{k!} \implies k! > \left(\dfrac{k}e\right)^k$$
Hence, we have
$$\dfrac{k}e < (k!)^{1/k} < k$$
Now conclude.