The matrix exponential is smooth

It is indeed a smooth function. A direct inductive argument can be given which is analogous (but a bit more tricky) to the argument that a complex function defined by a power series is smooth in the domain of convergence. In fact, it is actually easier to prove a more general statement and then apply it to deduce that $\exp$ is smooth.

Let $(A, \cdot, \| \cdot \|)$ be a finite dimensional complex Banach algebra. The reason we want to generalize our discussion to an arbitrary Banach algebra and not work only with $A = M_n(\mathbb{C})$ is that it makes the inductive argument easier. Show first that the power maps $p_k \colon A \rightarrow A$ given by $p_k(X) = X^k$ are continuously differentiable with differential given by

$$ dp_k|_{X}(Y) = X^{k-1}Y + X^{k-2}YX + \dots + XYX^{k-2} + YX^{k-1}.$$

The differential has this "strange" form because you don't know if $X$ and $Y$ commute. If they do, the formula above reduces to the usual formula $dp_k|_{X}(Y) = k X^{k-1} Y$.

Then show the following lemma:

Lemma: Let $(c_k)_{k=0}^{\infty}$ be a sequence of complex numbers such that $\sum_{k=0}^{\infty} c_k z^k$ converges on $B_{\mathbb{C}}(0,r)$. Define $f \colon B_A(0,r) \rightarrow A$ by $f(X) = \sum_{k=0}^{\infty} c_k X^k$. Then $f$ is well-defined and continuously differentiable. The differential of $f$ is given by

$$ df|_{X}(Y) = \sum_{k=0}^{\infty} c_k dp_k|_{X}(Y). $$

Finally, you can use the lemma inductively to deduce that $f$ is actually smooth and not just $C^1$. For full details, see Chapter 3 of the book "Structure and Geometry of Lie Groups" by Joachim Hilgert and Karl-Hermann Neeb.