Brazilian Mathematical Olympiad

If $k < 1$, then $k^n < 1$, so $n^{1/n} \ge 1$. We have $2^1 > 1$, and $2^{n+1} - 2^n = 2^n > 1 = (n+1) - n$, so a trivial induction shows that $2^n > n$. Hence $2 > n^{1/n}$.

Note that $3^2 > 2^3$, so taking 6th roots, $3^{1/3} > \sqrt{2}$. Note also that $4^{1/4} = \sqrt{2}$.

We show that $n^{1/n} < \sqrt{2}$ for $n > 4$. This is another trivial induction. We have $2^{5/2} = 4\sqrt{2} > 5$ and $2^{6/2} = 8 > 6$, so it is true for $n = 5$ and $n = 6$.