On the composition of formal power series I an attempt to compute the coefficients of the composition $f(g(x))$ of two power series $f(x) = \frac{1}{1-x}$ and $g(x) = \frac{1}{1-x}-1$, I used the definition of composition to get to 
$$f(g(x)) = \sum_{n=0}^{\infty} \left(\frac{1}{1-x} - 1\right)^n = \sum_{n=0}^{\infty} \left(\frac{x}{1-x}\right)^n$$
but I don't seem to be able to find a way to rewrite this in a way I can get the coefficients out, i.e. something of the form $\sum_{n=0}^\infty c_n x^n$.
Another way to compute the coefficients I know of, could be to use that $f(x) = \sum_{n=0}^\infty x^n$ and $g(x) = \sum_{n=1}^\infty x^n$ and actually compute each $c_N(x) = \sum_{n=0}^N f_n (g(x))^n = \sum_{n=0}^N (g(x))^n$, but this seems cumbersome if you need a lot of coefficients.
Using the transfer principle, on the other hand, one easily can get to 
$$ f(g(x)) = \frac{1}{1-\frac{1}{1-x} + 1} = \frac{1-x}{1-2x}$$
with expansion $1 + \sum_{n=1}^\infty 2^{n-1} x^n$.
Am I missing something obvious here or is it just not that easy to find the coefficients of the composition of two power series without leaving the "formal world"? I am wondering, because I personally am not the greatest fan of analysis and prefer to think in the formal world

We have $$f(g(x)) = \sum_{k = 0}^\infty \left( \dfrac x {1 - x} \right)^k .$$
Now, for $k \ge 1$, using $\dfrac 1 {(1 - x)^k} = \sum\limits_{m = 0}^\infty \binom{m + k- 1}{m} x^m$, we get
\begin{align*}
f(g(x)) & = 1 + \sum_{k=1}^\infty x^k \sum_m \binom{m + k - 1}{m} x^m\\
& = 1 + \sum_{k=1}^{\infty} \sum_m \binom{m + k - 1}{m} x^{m +k}.
\end{align*}
To collect together the coefficients of the same power of $x$ in the above summation, let $m + k = n$, to get
\begin{align*}
f(g(x)) & = 1 + \sum_{n=1}^\infty \sum_{m=0}^{n-1} \binom{n - 1}{m} x^n\\
& = 1 + \sum_{n = 1}^\infty 2^{n-1} x^n,
\end{align*}
since $\sum\limits_{m=0}^{n-1} \binom{n - 1}{m} = 2^{n-1}$.
