Let $M$ be a smooth oriented manifold, and let $M^E$ be an exotic copy, i.e homeomorphic but not diffeomorphic to $M$.

Is it true that $M\times M$ is diffeomorphic to $M\times M^E$?

I am interested in knowing the answer for closed manifolds.

One example which I can think of is an exotic $\mathbb R^4$. But we know that $\mathbb R^8$ has unique smooth structure. So in that case this is true.

One of the motivation for asking this question is following: I want to see an example of

symplectic manifold $\times$ non-symplectic manifold is symplectic manifold

. There are some manifolds whose some exotic structure doesn't admit symplectic structure. Also we know that Cartesian product of two symplectic manifold always admits symplectic structure. If the answer of my original question is positive, then that will imply the example I am looking for.

**EDIT:** A potential example of a symplectic manifold with non-symplectic exotic copy could be: Consider $E$ as elliptic surface. $F$ be fiber. Let $N_1= F\times D^2$. Consider $K$non-fibered knot is $S^3$ with non-monic $\Delta_k$. $T= K\times S^1$. And let $N_2=T\times D^2$ $\subset S^3\times S^1$. Them $E^E = (E-N_1)\cup (S3\times S^1-N_2)$ (gluing along meridian, sometimes it is called knot surgery). Then observe that in homology level we are not changing anything. So Freedman implies that it is homeomorphic with $E$. But Siberg-Witten invariant depends on Alexander polynomial of the knot. So thus we can conclude it has no symplectic structure. Similar ideas are written here https://arxiv.org/pdf/dg-ga/9612014.pdf