It's kind of self-evident. Any power of 21 will have a 1 in the units place, and adding 9 to it makes it 0 and hence the number is divisible by 5.
To prove the claim by induction, first establish the base case. I assume [tex]n\in\Bbb N[/tex], so for [tex]n=1[/tex] we have
[tex]21^1 + 9 = 21 + 9 = 30[/tex]
and of course 30 = 5×6 is divisible by 5.
Assume the claim is true for [tex]n=k[/tex], that [tex]5 \mid 21^k + 9[/tex]. This means that for some integer [tex]\ell[/tex], we can write
[tex]21^k + 9 = 5\ell[/tex]
Now the induction step: when [tex]n=k+1[/tex], we have
[tex]21^{k+1} + 9 = \bigg(21\times(21^k + 9) - 9\times21\bigg) + 9 \\\\ ~~~~~~~~~~~~~~ = 21\times5\ell - 9\times20 \\\\ ~~~~~~~~~~~~~~= 5\times(21\ell - 9\times4)[/tex]
which is divisible by 5. QED
