Let's determine the digits $A, B, C, D$ if

$$
A^{\overline{A B}}=\overline{C C B B D D C A}
$$
Obviously $A>1$; furthermore $A\left(2^{7}\right)^{4}=128^{4}>\left(10^{2}\right)^{4}$, so it must be at least 9 digits, thus the exponent can only be 25, and $B$ can only be 5.

Indeed, $2^{25}=\left(2^{10}\right)^{2} \cdot 2^{5}=1024^{2} \cdot 32=33554432$ satisfies the repetition of digits, so $C=3$, and $D=4$.