V OBM

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 $n=6$.