jmc

Look at the coefficients of the powers of 101: 1, -4, 6, -4, 1. You might recognize these as $\left(\genfrac{}{}{0ex}{}{4}{0}\right)$, $-\left(\genfrac{}{}{0ex}{}{4}{1}\right)$, $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$, $-\left(\genfrac{}{}{0ex}{}{4}{3}\right)$, $\left(\genfrac{}{}{0ex}{}{4}{4}\right)$. This suggests that the Binomial Theorem is in play. Indeed, we have

$$\begin{array}{rl}(101+(-1){)}^4& =\left(\genfrac{}{}{0ex}{}{4}{0}\right)\cdot 101^4\cdot (-1{)}^0+(\genfrac{}{}{0ex}{}{4}{1})\cdot 101^3\cdot (-1{)}^1+\\ & \left(\genfrac{}{}{0ex}{}{4}{2}\right)\cdot 101^2\cdot (-1{)}^2+(\genfrac{}{}{0ex}{}{4}{3})\cdot 101^1\cdot (-1{)}^3+\\ & \left(\genfrac{}{}{0ex}{}{4}{4}\right)\cdot 101^0\cdot (-1{)}^4\\ & =101^4-4\cdot 101^3+6\cdot 101^2-4\cdot 101+1.\end{array}$$

Therefore, we have $(101+(-1){)}^4=100^4=\boxed{100000000}$.