I am trying to prove the following inequality:
$$\left(\frac{e^n-1}{n}\right)^{2n+1} \leq \frac{(e-1)(e^2-1)...(e^{2n}-1)}{(2n)!}$$
For the base case $n=1$
$$0 \leq e-1$$
Lets assume the inequality holds for $n$, then I have to prove that:
$$\left(\frac{e^{n+1}-1}{n+1}\right)^{2(n+1)+1} \leq \frac{(e-1)(e^2-1)...(e^{2(n+1)}-1)}{(2(n+1))!}$$
Lets start with the RHS:
$$\frac{(e-1)(e^2-1)...(e^{2(n+1)}-1)}{(2(n+1))!}=\frac{(e-1)(e^2-1)...(e^{2n}-1)}{(2n)!} \frac{(e^{2n+1}-1)(e^{2n+2}-1)}{(2n+1)(2n+2)} $$
Now the LHS:
$$\left(\frac{e^{n+1}-1}{n+1}\right)^{2(n+1)+1}=\left(\frac{e^{n+1}-1}{n+1}\right)^{2n+1}\left(\frac{e^{n+1}-1}{n+1}\right)^{2}\left(\frac{e^{n}e-1}{n+1}\right)^{2n+1}\left(\frac{e^{n}e-1}{n+1}\right)^{2}$$
Well thats how far I generally got.
Is there a good way to rewrite the LHS, to solve this? Or is this approach wrong in general?
Looking at this inequality I also tried to use the convexity of $exp$ as an approach but this didn't lead anywhere.
By convexity, assuming $\frac{e^x-1}x=1$ at $x=0$, we have
$$\frac1{2n+1}\sum_{k=0}^{2n}\log\left(\frac{e^k-1}k\right)\ge \log\left(\frac{e^{\frac1{2n+1}\sum_{k=0}^{2n}k}-1}{\frac1{2n+1}\sum_{k=0}^{2n}k}\right)=\log\left(\frac{e^n-1}n\right)$$
which implies
$$\prod_{k=1}^{2n}\frac{e^k-1}k \ge \left(\frac{e^n-1}n\right)^{2n+1}$$
For the convexity we need to show that
$$f(x)=\log\left(\frac{e^x-1}x\right) \implies f''(x)=\frac{e^{2x}+1-e^x(x^2+2)}{x^2(e^x-1)^2}\ge 0$$
which is not much difficult using derivatives to study the sign for numerator.