I'm trying to prove the well-known identity $$\sum_k {n \choose k} = 2^n$$ with exponential generating functions (egf's). The idea is to note that the egf of the left hand side and the right hand side are the same. (Note that I don't actually care about proving this identity. It's the conceptual issue below that I care about.)
Here is my proof:
The left hand side is the binomial transform of the constant sequence $a_n = 1$, so its egf is $e^x e^x$. The right hand side has egf $$\sum_{k \geq 0} \frac{2^k}{k!} x^k = e^{2x}.$$ Since $e^x e^x = e^{2x}$, we're done.
But this confuses me. When I write $e^x e^x$ on the left hand side, I understand that to mean the product of two generating functions. When I write $e^{2x}$ on the right hand side, I understand it to mean the egf of the sequence $a_n = 2^n$. As analytic functions I know that these two are equal, but as generating functions I do not. That's exactly what I want to prove.
Can this proof be made valid? Specifically, can it be made valid without referencing analytic properties or the binomial theorem?
If you like, we're using a little theorem from complex analysis, that the only convergent power series for the zero function is $0$. It follows that if two convergent power series are equal as functions in a neighborhood of the origin, then their coefficients are termwise equal.