smc

We can use Legendre's Formula to find the number of $0$s in base $10$ $$\lfloor \frac{15}{5}\rfloor +\lfloor \frac{15}{25}\rfloor =3$$ So $h=3$. Likewise, we are looking for the number of $2^2$s and $3$s that divide $15!$, so we use Legendre's again. $$\lfloor \frac{15}{2}\rfloor +\lfloor \frac{15}{4}\rfloor +\lfloor \frac{15}{8}\rfloor =7+3+1=11$$ $$\lfloor \frac{15}{3}\rfloor +\lfloor \frac{15}{9}\rfloor =5+1=6$$ Thus, $3^6|15!$ and $2^{11}|15!\Rrightarrow (2^2{)}^5|15!$ So $k=5$, and $5+3=8$, which corresponds to answer $\boxed{8}$.