The problem is as follows:
'Let $f$ be a formal power series with $f_0=1$. Show that there is a unique formal power series $g$ with $g_0=1$ such that $g^d=f$, for some positive integer d.'
I am sure the answer here relies on the idea of unique fps inverses (i.e. we should be able to write the equality as $(g^d)(f)^{-1}=1$, where $1$ is the identity), but I can't see the approach from here.
Any hints or ideas would be much appreciated.
Assume characteristic zero, and $\,d\ne 0.\,$ Given $\,f\,$ as a formal power series $$ f := 1 + a_1 x + a_2 x^2 + \cdots $$ and suppose $\,f = g^d.\,$ Differentiate to get $$ D_x[g] = D_x[f]\,\frac{g}{d f}. $$ Integrate and use $\,g_0=1\,$ to get a recursion $$ g = 1 + \int D_x[f]\,\frac{g}{d f} dx. $$ This recursion shows existence and uniqueness of $\,g\,$ given $\,f.$ In more detail, define a sequence of functions by recursion $$ g_0 := 0+O(x),\quad g_{n+1} := \!1\!+\!\int D_x[f]\, \frac{g_n}{d f} dx. $$ The next two terms are $$ g_1\!=\!1 \!+\! \frac{a_1}d x \!+\! O(x^2), $$ $$ g_2\!=\! 1 \!+\!\frac{a_1}d x\!+\!\frac{ a_1^2(1\!-\!d)\!+\!2a_2d}{2d^2}\!+\!O(x^3). $$ This sequence converges to $\,g\,$ and any $\,g\,$ that satisfies $\,f=g^d\,$ agrees with $\,g_0\,$ and by recursion all of the $\,g_n.\,$