Solutions

By induction, let's prove that $(f\circ g)^{(n)}(a)$ is a sum of terms, each of which is a product of terms of the form

$$c(f^{(k)}\circ g)(a)(g^{(1)}(a))^{d_1}(g^{(2)}(a))^{d_2}\cdots(g^{(n)}(a))^{d_n},$$

with $c$, $k$, and $d_1,\ldots,d_n$ being nonnegative integers and $k,d_1,\ldots,d_n\le n$.

This is so for the base case $n=1$ since

$$(f\circ g)'(a)=(f'\circ g)(a)g'(a),$$

in which case the only term has $c=1$, $k=1$, and $d_1=1$.

For the induction step, suppose the conjecture holds for positive integer $n$ and that both $f^{(n+1)}(g(a))$ and $g^{(n+1)}(a)$ exist. Then $(f\circ g)^{(n+1)}(a)$ is the derivative of a sum of terms previously described. Hence, $(f\circ g)^{(n+1)}(a)$ is a sum of terms of the form

$$c(f^{(k+1)}\circ g)(a)(g')^{d_1+1}(g'')^{d_2}\cdots(g^{(n)})^{d_n}$$

or

$$c\cdot d_i(f^{(k)}\circ g)(a)(g')^{d_1}\cdots(g^{(i)})^{d_i-1}(g^{(i+1)})\cdots(g^{(n)})^{d_n},$$

with $i\le n$. Since $(f^{(k+1)}\circ g)(a)$ and $g^{(n+1)}(a)$ are defined, all constituents of the forms above are defined. Hence, $(f\circ g)^{(n+1)}(a)$ also is defined.