Two series relations, each one implies the other - from Andrews' partition book That's my first question here, and i was encouraged to post because my question in MathOverflow (HERE) was beautifully and fast answered. But my questions in not at research level...
As i said there, i'm working on a monograph about partitions, and one topic covered is the Simon Newcomb's problem. My main guide is the amazing "The Theory of Partitions", by George Andrews. I had two problems in understand some proofs on the book: one was solved by my question at MO, and the other is explained below.
The following lemma is in Andrews' book...

Lemma.
Let $r$ be an integer, and let $a_1, a_2, a_3, \ldots, b_1, b_2, b_3, \ldots$ be any numbers. Each of the following relations implies the other:
$
\begin{align}
\label{first_eq} \tag{1}
&a_n = \sum_{j=0}^{n-1}\binom{r - n + j}{j}b_{n-j}, \quad \forall n\geq 1;\\
\label{second_eq} \tag{2}
&b_n = \sum_{j=0}^{n-1}\binom{r - n + j}{j}(-1)^{j}a_{n-j}, \quad \forall n\geq 1.
\end{align}
$

I'm asking for an elementary, self-contained proof with no use of Chu-Vandermonde summation. If it uses generating functions, better, but binomial identities would be great too.
An obvious hint by Andrews is that one can just proof that (2) implies (1), because once it was done, a simple "variable change" proves the reverse implication, i.e., just consider $b'_{n} = (-1)^{n}b_n$ and $a'_{n} = (-1)^{n}a_n$.
I'm sorry for this basic question, and thanks in advance for the attention!
 A: Consider the formal series
$$
\begin{array}{}f(x)=\sum_{n=1}^\infty a_nx^n&\text{and}&g(x)=\sum_{n=1}^\infty b_nx^n\end{array}
$$
Then \eqref{first_eq} is equivalent to
$$
\begin{align}
f(x)
&=\sum_{n=1}^\infty\sum_{j=0}^{n-1}\binom{r-n+j}{j}b_{n-j}x^n\\
&=\sum_{n=1}^\infty\sum_{j=1}^{n}\binom{r-j}{n-j}b_jx^n\\
&=\sum_{j=1}^\infty\sum_{n=j}^{\infty}\binom{r-j}{n-j}b_jx^n\\
&=\sum_{j=1}^\infty\sum_{n=0}^{\infty}\binom{r-j}{n}b_jx^{n+j}\\
&=\sum_{j=1}^\infty b_j(1+x)^{r-j}x^{j}\\
&=(1+x)^rg\left(\frac{x}{1+x}\right)
\label{a}\tag{a}
\end{align}
$$
and \eqref{second_eq} is equivalent to
$$
\begin{align}
g(x)
&=\sum_{n=1}^\infty\sum_{j=0}^{n-1}\binom{r-n+j}{j}(-1)^ja_{n-j}x^n\\
&=\sum_{n=1}^\infty\sum_{j=1}^{n}\binom{r-j}{n-j}(-1)^{n-j}a_jx^n\\
&=\sum_{j=1}^\infty\sum_{n=j}^{\infty}\binom{r-j}{n-j}(-1)^{n-j}a_jx^n\\
&=\sum_{j=1}^\infty\sum_{n=0}^{\infty}\binom{r-j}{n}(-1)^na_jx^{n+j}\\
&=\sum_{j=1}^\infty a_j(1-x)^{r-j}x^{j}\\
&=(1-x)^rf\left(\frac{x}{1-x}\right)
\label{b}\tag{b}
\end{align}
$$
Simple algebraic substitutions yield that \eqref{a} and \eqref{b} are equivalent. Thus, \eqref{first_eq} and \eqref{second_eq} are equivalent.
A: As noted in the comments, this is a variation of the binomial inversion formula $\sum_{k=m}^n (-1)^{k-m} \binom{k}{m} \binom{n}{k} = \delta_{mn}$.  For proofs of this formula (which I will use), see this math.SE question.  It can also be proved easily using the trinomial revision formula I cite below.  From another perspective, binomial inversion says that the Pascal matrix is (up to sign) its own inverse.  For more on that see "Combinatorial interpretation of binomial inversion" and my answer to "Stirling numbers and inverse matrices." 
For your problem, I'll just prove that \eqref{second_eq} $\Rightarrow$ \eqref{first_eq}, since, as you note, that is sufficient.
First, reindex \eqref{first_eq} and \eqref{second_eq} to get the simpler forms $$a_n = \sum_{j=1}^n \binom{r-j}{n-j} b_j, \tag{1}$$ $$b_n = \sum_{j=1}^n \binom{r-j}{n-j} (-1)^{n-j} a_j. \tag{2}$$
Assuming that (2) is true, and starting with the right-hand side of (1), we have 
$$\sum_{j=1}^n \binom{r-j}{n-j} b_j = \sum_{j=1}^n \binom{r-j}{n-j} \sum_{k=1}^j \binom{r-k}{j-k} (-1)^{j-k} a_k = \sum_{k=1}^n  \sum_{j=k}^n \binom{r-j}{n-j} \binom{r-k}{j-k} (-1)^{j-k} a_k.$$
Now, apply the trinomial revision formula $\binom{r}{m} \binom{m}{k} = \binom{r}{k} \binom{r-k}{m-k}$.  (This is easy to prove; just write out the binomial coefficients in factorial form and regroup.  For a reference, see Concrete Mathematics, 2nd edition, p. 168.)  We get
$$\sum_{k=1}^n  \sum_{j=k}^n \frac{\binom{r}{n} \binom{n}{j}}{\binom{r}{j}} \frac{\binom{r}{j} \binom{j}{k}}{\binom{r}{k}} (-1)^{j-k} a_k = \sum_{k=1}^n \frac{\binom{r}{n}}{\binom{r}{k}} a_k \sum_{j=k}^n  \binom{n}{j} \binom{j}{k} (-1)^{j-k} .$$
Then, binomial inversion yields
$$=\sum_{k=1}^n \frac{\binom{r}{n}}{\binom{r}{k}} a_k \delta_{nk} = \frac{\binom{r}{n}}{\binom{r}{n}} a_n = a_n.$$
A: It's pretty easy to brute-force this proof, just by substitution and two basic identities.  It's actually true for arbitrary $r\in\mathbb{R}$, as long as you define $z\choose k$ for arbitrary real $z$ and non-negative integers $k$.
Given (1), replace the $a_{n-j}$ in the right hand side of (2) to get that you want:
$$b_n = \sum_{j=0}^{n-1} \sum_{k=0}^{n-j-1} (-1)^j {{r-n+j}\choose j}{{r-n+j+k}\choose k} b_{n-j-k}$$
Gathering terms, the coefficient of $b_{n-m}$ in the right hand side is (noting that j+k=m, or that k=m-j):
$$\sum_{j=0}^{m} (-1)^j{{r-n+j}\choose j}{r-n+m \choose m-j}$$
Note this is a function of $r-n$, so we can write:
$$f_m(z) = \sum_{j=0}^m (-1)^j{z+j \choose j}{z+m \choose m-j}$$
So, the right hand side of (2), after substitution, is:
$$\sum_{m=0}^{n-1} f_m(n-r) b_{n-m}$$
But ${z+m \choose m-j}{z+j \choose j} = {z+m \choose m} {m \choose j}$.  You can see this algebraically pretty easily, or you can prove it combinatorially for non-negative integers $z$, and then use that both sides are polynomials of degree $m$ in $z$, so they must agree everywhere.
So 
$$f_m(z) = {z+m \choose m} \sum_{j=0}^m (-1)^j {m \choose j} = {z+m \choose m} (1+(-1))^m$$
And therefore $f_m(z)=0$ if $m>0$, and $f_0(z)=1$.
But that means that the right hand side of (2) is equal to $b_n$.
