Find the last two digits of $7^{100}-3^{100}$

Alternatively, notice:$$(10-3)^{100}-3^{100} = \sum_{k=0}^{99}\binom{100}{k}10^{100-k}(-3)^k \equiv0 \pmod {100}$$

Hint:

The Carmichael function of $100$ is $20$,

so if $\gcd(a,100)=1$ then $a^{20}\equiv1\bmod100$,

so $a^{100}=(a^{20})^5\equiv1\bmod 100$.