# zero matrix to the power of 0

Why $0^0=I$?

I'd tried prove that considering $N^0$ where N is a Nilpotent matrix and then using the Cayley -Hamilton theorem

The same caveats as for the exponential of real numbers apply here. If exponentiation means repeated multiplication, then $A^0 = I$ is the base case for all $A$. Exponentiation by a continuous real parameter, on the other hand, should insist that $\mathbf{0}^0$ is undefined.
Just like the empty sum is defined to be the additive identity $0$, the empty product is usually defined to be the multiplicative identity $1$. For matrices, the multiplicative identity is the identity matrix.