$99 = 100 - 1 \rightarrow 99^5 = (100-1)^5$. We can now use Binomial Theorem to evaluate this.
$(100-1)^5 = ^5 \large C$$_0\;100^{5-1}1^0 -.^5 \large C$$_1\;100^{5-1}1^1 +.^5 \large C$$_2\;100^{5-2}1^2 -.^5 \large C$$_3\;100^{5-3}1^3 + ^5 \large C$$_5\;100^{5-5}1^5 $
$\qquad = 100^5 - 5(100^4) + 10(100^3) - 10(100^2) + 5(100) - 1$
$\qquad = 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1$
$\qquad = 9509900499$