A functional series for the exponential function 
Show that there is a sequence $(a_{n})$ such that $$e^z=1+\sum_{n=1}^{\infty}a_{n}{\sin(z^n)} ,|z|\le1 ,z$$ is a complex number.

We know that many functions as $f$ can be represented as $$f(x)=\sum_{n=1}^{\infty}a_{n}{\sin(nx)}$$ I consider $x^n$ instead of $nx$.
I am searching functions as $f$ and $g$ such that :$$g(z)=\sum_{n=1}^{\infty}a_{n}{f(z^n)}$$
 A: We are given the power series
$$ f(z)=\sum_{k=1}^\infty b_k z^k$$
where $\,b_1\ne 0\,$ and
$$ g(z)=\sum_{n=1}^\infty c_n z^n=
\sum_{j=1}^\infty a_jf(z^j)$$
which is always possible since the
$\,\{f(z^j)\}\,$ have leading term
$\,b_1z^j\,$.
Now, $$g(z)=\sum_{j=1}^\infty a_j\left(\sum_{k=1}^\infty b_k z^{jk}\right)=
\sum_{n=1}^\infty \left(\sum_{n=jk}a_jb_k\right)z^n.$$
Thus, $\,c_n = \sum_{n=jk}a_jb_k.\,$ Given
$\,\{b_n\}\,$ and $\,\{c_n\}\,$ we need to
solve for $\,\{a_n\}.\,$ For example,
$$ a_1 = \frac{c_1}{b_1},\;\;
a_2 = \frac{b_1c_2-b_2c_1}{b_1^2},\;\;
a_3 = \frac{b_1c_3-b_3c_1}{b_1^2}.$$
It seems that the solution for $\,a_n\,$ has
denominator $\,b_1^{A073093(n)}\,$ where
OEIS sequence A073093 is "Number of prime power divisors of n". The numerator is a linear combination of
$\,c_d\,$ where $\,d\,$ divides $\,n\,$
multiplied by powers of $\,b_k\,$ with an
integer coefficient. I don't find a simple
pattern.
In your question, $\,g(z)=e^z-1\,$ and $\,f(z)=\sin(z).\,$ We have
$$g(z)=\sum_{n=1}^\infty\frac1{n!}z^n\;\;
\text{ and }\;\;
f(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}
z^{2n+1}.$$
Solve for the $\,\{a_n\}\,$ coefficients to get
$$ \{a_n\}\!=\! 
\frac1{1!},\frac1{2!},\frac2{3!},\frac1{4!},
\frac0{5!},\frac{61}{6!},
\frac2{7!},\frac1{8!},\frac{20160}{9!},
\frac{-15119}{10!},\dots $$
and I don't find any simple pattern.
