Show that $n!^{1\over n}$ is increasing

Question

How can I show that

$$(n+1)!^{1\over n+1}>n!^{1\over n}$$

I hope to use this fact, along with the fact that $n!^{1\over n}$ is unbounded, to show that $\lim_\limits{n\rightarrow\infty}n!^{1\over n}=\infty$.

Answer 1

We have \begin{align*} n!^{n+1} &= n! \cdot n!^n \\ &= \underbrace{1 \cdot 2 \cdot \dotsb \cdot n}_{\text{$n$ factors}} \cdot n!^n \\ &< \underbrace{(n+1) \cdot (n+1) \cdot \dotsb \cdot (n+1)}_{\text{$n$ factors}} \cdot n!^n \\ &= (n+1)^n \cdot n!^n \\ &= (n+1)!^n \end{align*} and therefore $n!^{1/n} < (n+1)!^{1/(n+1)}$, as you wanted.

Answer 2

Raise both sides of your inequallity to the $n(n+1)$th power to get the equivalent inequality $$ (n+1)!^n > n!^{n+1}.$$ But $(n+1)! = (n+1)n!$ so this in turn is equivalent to $$ n!^n (n+1)^n > n!^{n+1}$$ and to $$(n+1)^n > n!$$ which obviously holds as $n!$ is the product of $n$ factors, each of which is smaller than $n$.

Answer 3

$$\left(\dfrac{(n+1)!}{n!}\right)^n=(n+1)^n>n!$$