Can Euler help you?

We note that finding the last two digits of $3333^{4444}$ can be obtained by reducing $3333^{4444}\pmod {100}$ . Since, $(3333,100) = 1$ , we can apply this theorem.

We first calculate that $\phi(100) =\phi(2^2)\phi(5^2)=(2)(5)(4)=40$ . Hence, it follows the Euler's theorem that $3333^{40}\equiv 1 \pmod{100}$ . Now let's apply the division algorithm on 4444 and 40 as follows :
$4444=40(111)+4$

Hence, it follows that: $3333^{4444} \equiv (3333^{40})^{111}. 3333^4 \equiv (1)^{111}.3333^4 \pmod{100} \equiv 33^4 \equiv 1185921 \equiv 21 \pmod{100}$

Hence the last two digits are 2 and 1.

$3333^{4444}=3^{4444}\cdot 1111^{4444} \\ (3^{4})^{1111}\cdot 1111^{4444}=81^{1111}\cdot 1111^{4444} \\ \text{Now finding last 2 digits : }81\cdot 41=33\boxed{21}$

How I found last 2 digits ?