Problem statement:
Suppose that $P(x) = a_1x + a_2x^2 + ··· + a_nx^n$ is a polynomial with integer coefficients, with $a_1$ odd. Suppose that $e^{P(x)} = b_0 +b_1x+b_2x^2 +···$ for all $x$. Prove that $b_k$ is nonzero for all $k ≥ 0$
My solution:
Let $f(x)=e^{P(x)}=\sum_{r=0}^\infty \frac{f^{(r)}(0)x^r}{r!}$ where $f^{(r)}$ denotes the $r$th derivative of $f$. (This is the Taylor series expansion rather than the exponential series expansion).
Note that by differentiating $f(x)$ we find that $f^{(r)}(x)$ is of the form $Q_r(x)e^{P(x)}$ where $Q_r(x)$ is a polynomial formed from a combination of derivatives of $P(x)$.
For instance, $f^{(1)}(x)=P^\prime(x)e^{P(x)}\implies Q_1(x)=P^\prime(x)$
$f^{(2)}(x)=[P^{\prime\prime(x)}+(P^\prime(x))^2]e^{P(x)}\implies Q_2(x)=P^{\prime\prime(x)}+(P^\prime(x))^2$
and so on.
The $b_k$ coefficients which we are required to show to be nonzero are of the form
$\frac{f^{(r)}(0)}{r!}=\frac{Q_r(0)e^{P(0)}}{r!}$
(from the Taylor series expansion), and since $\frac{e^{P(0)}}{r!}\neq0$, it suffices to show that $Q_r(0)\neq0 \ \forall r$. We will show that $Q_r(0)$ can be expressed as a sum of even integers and a single odd integer, meaning it is odd an thus nonzero.
We proceed by induction. It is easy to see that $Q_1(0)=P^\prime(0)=a_1e^{P(0)}\neq0$.
For the inductive step, firstly note that
$P^{(r)}(0)=\begin{cases} r!a_r & r=1,2,...,n \\ 0 & r>n \end{cases}$
Thus, since we are given that $a_1$ is odd, and $r!$ contains a factor of $2$ for $r>1$, it follows that $P^{(r)}(0)$ is odd for $r=1$ and even for $r>1$.
Now the $Q(x)$ polynomials as we have seen can each be expressed as a sum of terms whose factors are the derivatives of $P(x)$. We claim that $Q_r(0)$ can be expressed (for all $r$) as a sum of terms all of which are even integers apart from a single power of $P^{(1)}(0)$ which is odd, and hence that the sum overall is odd.
$Q_1(x)=P^{(1)}(x)$ and so $Q_1(0)$ satisfies this requirement.
Suppose $Q_m(0)$ also satisfies the requirement.
We differentiate $f^{(r)}(x)$ to find that
$Q_{m+1}(x)=Q_m^\prime(x)+P^\prime(x)Q_m(x)$
During differentiation, the single term in $Q_m^\prime(x)$ which was initially a power of $P^{(1)}(x)$ acquires a factor of $P^{(2)}(x)$, and thus becomes even within the term $Q_m^\prime(x)$. On the other hand, within the term $P^\prime(x)Q_m(x)$, the single odd term simply increases its power by one, and thus remains odd.
Hence, by induction, $Q_m(0)$ has a unique odd integer term and is thus nonzero for all $m$. This completes the proof.
One concern I have is whether it's ok to use Taylor series in this way to express a function that was already a different kind of infinite series? (I haven't done this before). Also just in general wondering whether there is anything else about the proof that's insufficient (though if I'm not allowed to ask that here - the rules aren't exactly clear - please do tell me and I'll go somewhere else).