Last two digits of $3^{400}$

I am trying to calculate the last two digits of $3^{400}$, but working $\bmod {100}$ is proving intractable. I did the last digit case pretty handily, and now I'm stuck at last two.

We have $3^{400}=(-1+10)^{200}$. The first term in the expansion via the Binomial Theorem is $1$. The other terms are divisible by $100$. Indeed by $1000$.

Are you familiar with the generalization of Fermat's little theorem?

From it you know that $3^{\phi(100)}\equiv 1\pmod{100}$, where $\phi$ is the Euler totient function. I get $\phi(100)=40$, and so $3^{400}\equiv 1\pmod{100}$ if I haven't made any errors.

$$3^{400}=3^{5\cdot 2\cdot 2\cdot 20}\equiv 43^{2\cdot 2\cdot 20}\equiv 49^{2\cdot 20} \equiv 1^{20} \equiv 1 \quad(\bmod 100) $$