Are there exotic balls?
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.
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.