Are there exotic balls? Suppose $M$ is a compact smooth manifold with boundary, which is homeomorphic to the compact ball $\mathbb{B}^d\subset \mathbb{R}^d$. Must $M$ be diffeomorphic to $\mathbb{B}^d$ or are there exotic smooth structures?
I suspect that the answer is well known or might follow from a simple argument that experts (which I am certainly not) have up their sleeves.
A natural candidate in $d=4$ would be to take $M$ to be the compact unit-ball in an exotic $\mathbb{R}^4$, but I don't know how to check whether $M$ is exotic (or even whether it has a smooth boundary to be honest).
 A: In dimension 4, one has the following equivalence:

The set $\mathcal B_4$ of compact smooth manifolds homeomorphic to $B^4$, considered up to oriented diffeomorphism, is in canonical bijection with the set $\mathcal S_4$ of compact smooth manifolds homeomorphic to $S^4$, considered up to oriented diffeomorphism.

Proof. We construct maps both ways, which will be clearly inverse.
$C: \mathcal B_4 \to \mathcal S_4$. Pick $B \in \mathcal B_4$. By the resolution of the 3D Poincare conjecture, there is some oriented diffeomorphism $\varphi: \partial B \to S^3$. One may then "cap off the boundary": define $$C(B) = B \cup_\varphi B^4,$$ defined as $B \sqcup B^4$, modulo identifying $\partial B \cong S^3$ via the diffeomorphism $\varphi$.
$C$ is well-defined by Cerf's theorem that $\pi_0 \text{Diff}^+(S^3) = 1$: all diffeomorphisms are the same up to isotopy. Isotoping $\varphi$ above does not change the diffeomorphism type; so $C$ is a set map.
Conversely one has $D: \mathcal S_4 \to \mathcal B_4$, with $D(S)$ given by deleting the interior of some oriented embedding $\iota: B^4 \to S$. Oriented embeddings of balls into a connected manifold are unique up to isotopy (Palais, but straightforward); this is true in all dimensions.
Clearly $D(C(B)) = B$ (delete the ball you glued) and $C(D(S)) = S$ (glue in the ball you deleted). Therefore $C,D$ are inverse bijections.

However one has the following for $n \geq 6$.

The set $\mathcal B_n$ is trivial.

Proof: Pick $B \in \mathcal B_n$. Delete a standard ball from its interior; then we are provided with a compact manifold $W$, with $\partial W = S^{n-1} \sqcup \partial B$, so that $W \cup_{S^{n-1}} B^n = B$.
$W$ is an h-cobordism (algebra). Therefore by the h-cobordism theorem there is a diffeomorphism $W \cong S^{n-1} \times [0,1]$, which sends $S^{n-1}$ to $S^{n-1} \times \{0\}$ by the identity map. Therefore $W \cup_{S^{n-1}} B^n \cong B^n$. Therefore $B \cong B^n$.

Of course there are no exotic n-balls n = 1,2,3. For n=5 I do not know. I think the answer is that $\partial: \mathcal B_5 \to \mathcal S_4$ is an injection (maybe bijection?) but I don't know off the top of my head. Some googling or looking on Math Overflow should help. Obviously "exotic balls" will get you nowhere but some buzzwords like diffeomorphism might.
A: This is an addendum regarding dimensions 4 and 5: While the (smooth) h-cobordism argument for 5-dimensional cobordisms no longer works, Milnor proves in his "h-cobordism" book (pages 110-111, Proposition C) that if $M$ is a 5-dimensional smooth compact manifold homeomorphic to $D^5$, whose boundary is diffeomorphic to $S^4$, then $M$ is diffeomorphic to $D^5$. According to theorems of Milnor-Kervaire and Wall, if $M^4$ is an exotic $S^4$, then it bounds a smooth contractible 5-manifold $W$.  The manifold $W$ then will have to be homeomorphic to $D^5$ (by the topological h-cobordism theorem for 5-dimensional h-cobordisms, due to Freedman). Thus, the existence of an exotic $D^5$ is equivalent to the existence of an exotic $S^4$.
In dimension 4 one would need a bit more: If $W$ is a smooth 4-manifold homeomorphic to $D^4$, then its double $DW$ is homeomorphic to $S^4$. If $DW$ is diffeomorphic to $S^4$, assuming, in addition, smooth Schoenflies conjecture in dimension 4, one would obtain that $W$ is diffeomorphic to $D^4$. If smooth Schoenflies conjecture fails, then you would obtain a smooth submanifold inside $S^4$ which is homeomorphic but not diffeomorphic to $D^4$. (The same deal if the smooth Poincare conjecture fails in dimension 4.) So, "in essence," in dimension 4, the problem is equivalent to Poincare+Schoenflies.
