Answer
see below.
Work Step by Step
1. Test for $n=1$, $1+a\ge 1+a$, it is true.
2. Assume the statement is true for $n=k$, that is $(1+a)^k\ge1+ka$,
3. For $n=k+1$, with $a\gt0$, we have $(1+a)^{k+1}=(1+a)(1+a)^k\ge(1+a)(1+ka)=1+ka+a+ka^2=1+(k+1)a+ka^2\ge1+(k+1)a$, thus it is true for $n=k+1$,
4. We can conclude that the statement is true for any integer $n$.