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I'm trying to proof the following inequality $$(n+k)^k (n-k)! \leq (2n)^n, \quad \forall k = 0,...,n.$$

It's clear for $k = 0,1$ and $n$.

For $k = 0$, $(n+0)^0(n-0)! = n! \leq n^n \leq (2n)^n$.

For $k = 1$, $(n+1)(n-1)! = n(n-1)! + (n-1)! \leq n^n + n^n = 2n^n \leq (2n)^n$.

For $k = n$, $(2n)^k \, 0! = (2n)^k \leq (2n)^n$.

However, I didn't get to proof for an arbitrary $0 \leq k\leq n$.

Help?

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    @hkmather802 Not quite. The inequality $k^2\ge nk$ holds for $k=0$, $k=n$, and $k=n+1$ for all $n$, but not for all $k\in\{0,\ldots,n\}$. – 2017-02-08

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$$ (n+k)^k \le (2n)^k $$ and $$ (n-k)! \le (n-k)^{n-k} \le (2n)^{n-k} $$