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Is it possible to prove this inequality: $$(x+v-1)^{x-1} / {(x-1)!} \le x^v, \forall \ (x,v) \in \mathbb{N^2}$$

I tried to find a counterexample, but I couldn't and I have no clue of where to start for proving it.

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    $\begingroup$ I think $x=3$ and $v=1$ is a counterexample $\endgroup$ – Ákos Somogyi Apr 1 '18 at 15:06
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This inequality is not true. Setting $v=1$, we have $$\frac{x^x}{(x-1)!} \leq x \iff x^{x-1} \leq (x-1)!$$ Which is definitely not true. This inequality does not hold the other way either, since for $u=v=2$, we have $$(2+2-1)^1 \le 2^2 \iff 3 \le 4$$

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    $\begingroup$ As in the comment from Akos Somogyi, $3^{3-1}=9>2=(3-1)!$ $\endgroup$ – DanielWainfleet Apr 1 '18 at 21:11

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