Given two square matrices $A$ and $B$, is $(AB)^2 = A^2B^2$ true or false?

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I tried with a counter example and it came out that this claim is false.

I took a matrix
$$
A=
\left[ {\begin{array}{cc}
2 & 1\\
3 & 2\\
\end{array} } \right]
$$

and a matrix

$$
B=
\left[ {\begin{array}{cc}
1 & 3\\
4 & 1\\
\end{array} } \right].
$$

I first calculated $$AB$$

and I got:

$$
AB=
\left[ {\begin{array}{cc}
6 & 7\\
11 & 11\\
\end{array} } \right]
$$

and then I calculated $$AB \times AB$$ that is same as $$(AB)^2$$
and I got:
$$
(AB)^2=
\left[ {\begin{array}{cc}
85 & 119\\
187 & 198\\
\end{array} } \right]
$$

then I calculated $$A \times A$$ that is the same as $$A^2$$ and I got
$$
A^2=
\left[ {\begin{array}{cc}
7 & 4\\
12 & 13\\
\end{array} } \right]
$$

and after I calculated $$B^2$$

$$
B^2=
\left[ {\begin{array}{cc}
13 & 6\\
8 & 13\\
\end{array} } \right]
$$

and at the end I calculated $$A^2B^2$$ and I got:

$$
A^2B^2=
\left[ {\begin{array}{cc}
123 & 94\\
250 & 241\\
\end{array} } \right]
$$

so I am deducing that the claim at the beginning is false, so
$$(AB)^2 \neq A^2B^2$$

Is this right?