# Binomial expansion?

The next-to-last step in the proof on pg 1 of this article

makes this substitution

$$\sum_{r=0}^\infty \binom {m+r-2}{r} q^r = (1-q)^{1-m}$$

I don't see it. How does
$$(1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^r$$
or
$$(1-q)^{1-m} = \sum_{r=0}^\infty \binom{1-m}{r} (-q)^{1-m-r}$$
transform to the expression given by Haldane?

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