This follows easily from the identity $ x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \dots + y^{n-2} x + y^{n-1}) $ where $x = 1/(w-z)$ and $y=1/(w-a)$. I am writing this so that you know this very useful identity, which works in any commutative ring. To prove it, you can simply expand the factor on the right.
EDIT : I am putting it here explicitly because of J.M.'s comment and because I never actually noticed it could be shown in this way, but when you look at the identity $ \frac{1-z^n}{1-z} = z^{n-1} + \dots + z + 1 $ , you can let $z = y/x$ and see that $ 1-\left( \frac yx \right)^n = \left( 1 - \frac yx \right) \left( \left( \frac yx \right)^{n-1} + \dots + \frac yx + 1 \right). $ Multiplying by $x^n$ on the left hand side and by $x\cdot x^{n-1}$ on the right hand side, we also get the identity.