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A proof I did recently called upon a "fact" which my prof called without giving explanation or proof, which is the "fact" that $\sqrt[n]{n}>1$, how can this be shown?

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    $\begingroup$ Here: $n> 1^n = 1$. $\endgroup$ – user14717 Nov 7 '14 at 2:50
  • $\begingroup$ What context were you stating this in? $\endgroup$ – 9301293 Nov 7 '14 at 3:11
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You should give a more precise context. But, let's try something. If $n \in \mathbb N, n\ge1$, then $$n \ge 1^n \Leftrightarrow \sqrt[n]n\ge1\,.$$ PS: please note I used $\ge$ instead of the stricter $>$.

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Suppose $x>0$.

If $p>0$, then $x^p>1 \iff x>1$, and $x^p<1 \iff x<1$.

If $p<0$, then $x^p<1 \iff x>1$, and $x^p>1 \iff x<1$.

The second statement follows from the first by just taking reciprocals, which reverses the inequality.

The first statement is true because the function $f(p)= x^p$ is a strictly increasing function of $p$ for fixed $x>1$, and a strictly decreasing function of $p$ for fixed $x<1$.

In your case, take $p=1/x$ (a positive number) and you are considering $x=n$ for integral $n>1$ (so certainly $x>0$).

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