$ f(x) = x + x^2 + x^4 + x^8 + x^{16} + x^{32} + \dots$
Then coefficient of $x^{10}$ in $f(f(x))$ is?
Solution: We can write $f(f(x)) = f(x) + f(x)^2 + f(x)^4 + \dots$
Then, coefficient of $x^{10}$ in $f(x)= 0$ and the coefficient of $x^{10}$ in $(f(x))^2 = 2$ as $(2,8)$ and $(8,2)$ are the two exponent combinations but how to find coefficient of $x^{10}$ in $f(x)^4$ and so on?
As you noticed, $$ f(f(x)) = f(x)+f(x)^2+f(x)^4+f(x)^8+f(x)^{16}+\ldots $$ on the other hand $f(x)^m = x^m +o(x^{m})$, hence $$ [x^{10}]f(f(x)) = [x^{10}]\left( f(x)+ f(x)^2+f(x)^4 + f(x)^8 \right) $$ and $$[x^{10}]f(x)^m = \left|\left\{(a_1,\ldots,a_m): a_1+\ldots+a_m=10, a_k = 2^j\right\}\right|. $$ In particular $[x^{10}]f(x)=0$ (trivial by inspection), $[x^{10}]f(x)^2 = 2 $ since $10=2+8=8+2$, $[x^{10}]f(x)^4 = 10$ since $10=4+4+1+1=4+2+2+2$ and $[x^{10}]f(x)^8=28$ since $10=1+1+1+1+1+1+2+2$. The final outcome is $$ [x^{10}]f(f(x)) = 0+2+10+28 = \color{red}{40}.$$