$e^{0\times I}$ in Wolfram Alpha broken

Question

How comes that

$$
e^{0_2} = e^{\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}} = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} = 1_2
$$

according to Wolfram Alpha? Shouldn't $e$ to the power of a square matrix consisting of all zeros be the identity matrix of the same size? By my reasoning, we have

$$
e^{0_2}\times e^{0_2} = e^{0_2 + 0_2} = e^{0_2}
$$

(I know that $e^A\times e^B$ is not equal to $e^{A+B}$ in general, since $e^A$ and $e^B$ might not commute, but that is how I think of it in this case), but

$$
1_2\times 1_2 = \begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}2 & 2 \\ 2 & 2\end{pmatrix} = 2_2 \neq 1_2,
$$

so we must have

$$
e^{0_2} \neq 1_2.
$$

Is there something wrong with this reasoning?

Answer 1

If you double-check Wolfram, you'll see that you performed an elementwise operation:

If you instead use MatrixExp instead of e^, you will get the desired result (link):