Prove that $7^{100} - 3^{100}$ is divisible by $1000$

Wolfie A is never the right way.

By the Chinese remainder theorem, all you need is to prove both $$7^{100}\equiv3^{100}\pmod8$$ and $$7^{100}\equiv3^{100}\pmod{125}.$$ You have already done the latter. But $7^2\equiv1\pmod 8$ and $3^2\equiv1\pmod 8$ so it's a fair bet that $7^{100}\equiv3^{100} \pmod8$ too.

\begin{eqnarray} 7^{100}-3^{100} &=& (10-3)^{100}-3^{100}\\ &=& \underbrace{{100\choose 0}10^{100}-{100\choose 1}10^{99}\cdot 3+...-{100\choose 97}10^3 \cdot 3^{97}}_{10^3k}+{100\choose 98}10^2 \cdot 3^{98} -{100\choose 99}10 \cdot 3^{99}+3^{100}-3^{100}\\ &=&1000k +50\cdot 99\cdot10^2 \cdot 3^{98} -100\cdot 10 \cdot 3^{99} \end{eqnarray}

By the binomial theorem, $$ 3^{100} =(7-10)^{100} =7^{100}-\binom{100}{1}7^{99}10+\binom{100}{2}7^{98}10^2 + 10^3a $$ Now $ \binom{100}{1}10=1000 $ and $ \binom{100}{2}10^2 = 495000 $