Discuss the monotonicity of $\sqrt[n]{n!}$. It seems that $\sqrt[n]{n!}$ is increasing because it turns to $+\infty$ as $n\to+\infty$. How to prove it?
 A: It is enough to show that $\frac{1}{n}\log(n!)=\frac{1}{n}\sum_{k=1}^{n}\log(k)$ is increasing.
Since $\log(x)$ is concave on $\mathbb{R}^+$, this is a simple consequence of Karamata's inequality:
$$ \tfrac{1}{n}\cdot\log(1)+\tfrac{1}{n}\cdot\log(2)+\ldots+\tfrac{1}{n}\cdot\log(n)+0\cdot \log(n+1)\\ \leq \tfrac{1}{n+1}\cdot\log(1)+\tfrac{1}{n+1}\cdot\log(2)+\ldots+\tfrac{1}{n+1}\log(n+1).$$

As an alternative approach, $n!^{n+1}\leq (n+1)!^n$ is equivalent to $n!\leq (n+1)^n$ which is trivial.
A: Hint:
$$
\begin{align}
\frac{n!^{n+1}}{(n+1)!^n}
&=\frac{n!}{(n+1)^n}\\
&=\frac1{n+1}\cdot\frac2{n+1}\cdot\frac3{n+1}\cdots\frac{n}{n+1}
\end{align}
$$

Another Hint: Notice that $n!^{\frac1n}$ is the geometric mean of first $n$ natural numbers. $(n+1)!^{\frac1{n+1}}$ is the geometric mean of the same set of natural numbers and one larger natural number.
A: Let $a_n = \sqrt[n]{n!}$
Let's show that $a_{n+1}>a_n$.
Need to show: $\sqrt[n+1]{(n+1)!} > \sqrt[n]{n!}$
Take $n+1$ power on both sides.
$$(n+1)! > (n!)^{\frac{n+1}{n}}=(n!)^{1+1/n}$$
Divide both sides by $n!$
Then$$n+1 > (n!)^{1/n}$$
$$(n+1)^n > n!$$
Since
$$(n+1)(n+1)\cdots(n+1) > n\cdot (n-1)\cdots 1$$
(Each side has $n $ elements)
(Write backwards to complete the proof)
A: Suppose that for some $n$ we have $$(n+1)!^{\dfrac {1}{n+1}}\leq n!^{\dfrac {1}{n}}$$
Take both sides to the power of $n(n+1)$ to obtain $$(n+1)!^n\leq n!^{n+1}$$
From that we arrive at $$n!^n(n+1)^n\leq n!^nn!$$
And now we obtain $$n^n<(n+1)^n\leq n!$$ which is clearly false so we arrived at a contradiction, so your sequence is strictly increasing.
