prove-p-P-p-2pi-k-1-n-a-k-a-1-a-2-a-n-

Question Number 217681 by SdC355 last updated on 18/Mar/25

$$\mathrm{prove} \\ $$$$\underset{{p}\in\mathbb{P}} {\overset{\infty} {\prod}}\:{p}=\sqrt{\mathrm{2}\pi} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{a}_{{k}} ={a}_{\mathrm{1}} {a}_{\mathrm{2}} \centerdot\centerdot\centerdot{a}_{{n}} \\ $$

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