I am asking this question because I am unfamiliar with proofs and want to ask this question through an example:
If |x| < 1 then the coefficient of $x^n$ in $(1-x)^{-(N+1)}$ is $\begin{pmatrix} N + n\\ n \end{pmatrix}$. (Using the extended binomial theorem)
Which is equivalent to the coeffiicient of ${x^n}$ in $(1-x)^{-1} \cdot (1-x)^{-N}$. $(1-x)^{-1} \cdot (1-x)^{-N} = (1+x+x^2 + x^3...)\cdot (1-x)^{-N}$. The Coefficient of $x^n$ here is $\sum_{j = 0}^{ n}\begin{pmatrix} N + j - 1\\ j \end{pmatrix}$. Because you have the $x^n$ term from $(1-x)^{-N}$ multiplied by 1, the $x^{n-1}$ term multiplied by $x^2$... etc till the $x^0$ term multipled by $x^n$.
Therefore you can say that $\begin{pmatrix} N + n\\ n \end{pmatrix} = \sum_{j = 0}^{ n}\begin{pmatrix} N + j - 1\\ j \end{pmatrix}$.
To me it seems that the above statement is always true. However when proving it we assumed that |x| < 1. So are we allowed to prove statements using conditional statements. Or is it only allowed in this case because the final expression had no x in it?
You proved that the coefficient of $x^n$ for $(1-x)^{-(N+1)}$ can be expressed in two different forms when $|x|<1$.
If two power series have the same value in an interval they have both the same coefficients. So you proved the combinatorial identity because then also the coefficient of $x^n$ must be the same, the fact that $|x|<1$ has no influence in this case.