Is it right that repeated AES encryption under the same key will eventually lead back to the original plaintext?
The reasoning is that AES acts like a reversible permutation of the blocks, the composition of several permutations is yet another permutation (closure), and because the number of permutations is finite, it must eventually get back to where it started.
If I write $E \circ E = E^2$ to mean "double AES" and $E^N$ to mean $N$ applications of AES (under the same key), then I can state the hypothesis symbolically as $E^N = I$ for some $N$, where $I$ is the identity permutation.
If this is correct, then it seems that $N-1$ applications of AES behaves identically to the inverse/decryption function, $D$, because
$$E^N = E \circ E^{N-1} = E \circ D = I$$
Is this statement correct?