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How do we prove that $(x-1)!leq{(frac{x}{2})^{x-1}}$?

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0 $begingroup$ I plotted the graph of $y=(x-1)!$ and $y=(frac{x}{2})^{x-1}$ and found that the latter is always greater than the former for $x>2$ . Also, equality holds at $x=2$ . Would someone please help me to prove it mathematically? inequality share | cite | improve this question edited Jan 6 at 9:37 Shashwat1337 asked Jan 5 at 18:26 Shashwat1337 Shashwat1337 95 9 $endgroup$