Problem: Prove for natural numbers $n > 2$, $$(\sqrt{2!}-1)((3!)^{\frac{1}{3}}-\sqrt{2!})\cdots(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}) < \frac{n!}{(n+1)^n}.$$
I am unable to do this one. Please help.
My attempts: By AM-GM we get,
$(\sqrt{2!}-1)((3!)^{\frac{1}{3}}-\sqrt{2!})\cdots(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}) < \left( \frac{((n+1)!)^{\frac{1}{n+1}}-1}{n}\right)^{n} $.
So, now its enough to show, $$\left( \frac{((n+1)!)^{\frac{1}{n+1}}-1}{n}\right)^{n} < \frac{n!}{(n+1)^n}$$ But I don't know how to do that.
I also tried with induction, but it did not work
The last inequality can be written as $$(n+1)\cdot\left((n+1)!^{\frac{1}{n+1}}-1\right)<n\cdot n!^\frac{1}{n}$$ or: $$(n+1)\cdot (n+1)!^{\frac{1}{n+1}} -n\cdot n!^\frac{1}{n}< n+1$$ that is straightforward to prove through Stirling's approximation: $n!\approx\frac{n^n}{e^n}\sqrt{2\pi n}$ gives that the LHS behaves like $\frac{2}{e}\cdot(n+1)$ and $\frac{2}{e}<1$.