This is a bit late, but I thought of another example. Looking at the initial coefficients of the power series 
$$
\sqrt[3]{1+x} = 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 - \frac{10}{243}x^4 + \frac{22}{279}x^5 + \cdots
$$
we may guess that each coefficient has a 3-power denominator.  From Taylor's formula, the coefficient $c_k$ of $x^k$ is $$c_k = \frac{(1/3)(1/3-1)\cdots(1/3-k+1)}{k!}.$$
To prove the fraction $c_k$ has a 3-power denominator, we will show its denominator is not divisible by any prime $p \not= 3$ by viewing $1/3$ as a *$p$-adic limit* of positive integers: $1/3 = \lim_{r \rightarrow \infty} a_r$, where $a_r$ is the $r$th truncation of $1/3$ in its $p$-adic expansion. Since the polynomial $x(x-1)\cdots (x-k+1)/k!$ is continuous for the $p$-adic topology, $c_k$ is the limit of 
$a_r(a_r-1)\cdots(a_r-k+1)/k!$ as $r \rightarrow \infty$, and these terms are all integers, hence $p$-adic integers, so the limit is a $p$-adic integer, and thus $c_k$ has no $p$ in its denominator. 

For more discussion of this, see my answer at 
https://math.stackexchange.com/questions/136206/show-that-sqrt1t-lies-in-mathbbz1-2-t/136288#136288.


In a similar spirit, identities for formal power series (like the chain rule for $f(g(X))$ when $g(X)$ has constant term 0) may be easier to prove for polynomials using induction and then one can appeal to $X$-adic continuity to pass to the limit and get the same result for power series. Here we insert a power series into the family consisting of its truncated polynomial approximations along with the power series itself as the limit.