Prove that $(n!)^{\frac{1}{n}} < ((n+1)!)^{\frac{1}{n+1}}$ I have been tasked with proving that $(n!)^{\frac{1}{n}} < ((n+1)!)^{\frac{1}{n+1}}$ for every integer $n \geq 1$. My instinct is to use induction, but I have gotten stuck.
Base Case - $n=1$: $(1!)^{\frac{1}{1}} < ((1+1)!)^{\frac{1}{1+1}} \rightarrow 1 < 2^{\frac{1}{2}}$, which is of course true.
Inductive Hypothesis - $((n-1)!)^{\frac{1}{n-1}} < (n!)^{\frac{1}{n}}$
Inductive Proof - WTS $(n!)^{\frac{1}{n}} < ((n+1)!)^{\frac{1}{n+1}}$
I am trying to go about this by expanding $((n+1)!)^{\frac{1}{n+1}}$ to show that it must be greater than $(n!)^{\frac{1}{n}}$. Is this correct? It is where I have gotten stuck.
$((n+1)!)^{\frac{1}{n+1}}=((n+1)(n!))^{\frac{1}{n+1}} = (n+1)^{\frac{1}{n+1}} * (n!)^{\frac{1}{n+1}}$
I don't know where to go from here, or how to use my IH in this proof. I considered making my IH $(n!)^{\frac{1}{n}} < ((n+1)!)^{\frac{1}{n+1}}$, and trying to prove $((n+1)!)^{\frac{1}{n+1}} < ((n+2)!)^{\frac{1}{n+2}}$, but I got just as stuck in the proof.
 A: No need to use induction.
$$(n!)^{\frac{1}{n}}<((n+1)!)^{\frac{1}{n+1}}\impliedby n!<((n+1)!)^{\frac{n}{n+1}}\impliedby (n!)^{n+1}<((n+1)!)^n$$
The last inequality holds since for all $1\le i\le n$ and $i\in \mathbb{N}$ we have $i<(n+1)$. 
A: HINT: Rearranging the inequality to get rid of the fractional exponents is the easiest approach, but you can also look at the ratio of consecutive terms:
$$\left(\frac{(n+1)!^{1/(n+1)}}{n!^{1/n}}\right)^{n+1}=\frac{(n+1)!}{n!\cdot n!^{1/n}}=\frac{n+1}{n!^{1/n}}>\frac{n+1}{(n^n)^{1/n}}>\ldots$$
A: It is possible to proof this by induction, however I don't consider it nice and as shown by others, there really are nice solution:
Let's say the inequality holds for 1 to n:
So it is $ (n-1)! ^{1 \over n-1}   < n!^ {1 \over n} ; $
Now you want to proof the inequality for $ n+1; $
$ n! ^{1 \over n}   < (n+1)!^ {1 \over (n+1)} ; $ 
Now we are pushing the exponents around to make things look nicer:
$ n! ^ {n+1}   < (n+1)!^n $
Now we rewrite the inequality so that we can use that the inequality already holds for 1 to n: 
$ (n-1)! * (n-1)! ^ n * n^{n+1} < n ^{n-1} * n! * (n+1)^n $
Using what we already know, we get:
$ (n-1) ! * n^{n+1} < n! * (n+1)^n $
$ n^n < (n+1)^n $ - and this looks to be obviously true, thus the inequality is proven for all n.. 
