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Prove that the sequence $\sqrt[n]{n!}$ diverges to infinity [duplicate]

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Currently, I have that $\sqrt n \leq \sqrt[n]{n!}$ for all integers $n≥1$, and since the sequence in the lower bound $\lim\limits_{n\to \infty}\sqrt n=\infty $ blows up, so does the upper bound $x_n=\sqrt[n]{n!}$

This feels quite hand-wavey and unsatisfying to me, so I was curious if there any other creative / more formal ways to prove its divergence?


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