#### Answer

See below.

#### Work Step by Step

Proofs using mathematical induction consists of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number, then we prove that then the statement also holds for $n + 1$.
Hence here: 1) For $n=5: 2^5=32\gt5^2=25$
2) Assume for $n=k: 2^k\gt k^2$.
Then for $n=k+1$: $2^{k+1}\gt2k^2\gt k^2+2k+1=(k+1)^2$
because $k^2\gt2k+1$ for $k\geq5$ from Exercise 41$.
Thus we proved what we wanted to.