Find the remainder!

We can take $7^{700}$ (mod 4) and (mod 25), then we can use the Chinese Remainder Theorem to find $7^{700}$ (mod 100).

$7^{700}\equiv (-1)^{700}\equiv 1$ (mod 8) $7^{700}\equiv {49}^{350}\equiv (-1)^{350}\equiv 1$ (mod 25)

Using the Chinese remainder theorem, we get that $7^{700}\equiv \boxed{1}$ (mod 100).

$Mathematica$

PowerMod[7,700,100]