For any natural number $k, k^2<2^k$ if $5 \leq k$. [Hints. For this problem, freely assume the standard facts of arithmetic, including exponentiation and inequalities. If $k^2<2^k$, then $2 k^2<2^{k+1}$. Also, for $5 \leq k$, we have $k(k-2)>1$, from which one can eventually obtain $\left.2 k^2>k^2+2 k+1.\right]$