Stereographic projection in $\mathbb{R}^n$

I'm trying to verify that if $\phi(x_1, \cdots x_n, x_{n+1}) = \left(\frac{x_1}{1-x_{n+1}}, \cdots \frac{x_n}{1-x_{n+1}} \right)$ then

$$\phi^{-1}(\zeta_1 ,\cdots \zeta_n) = \left(\frac{2\zeta_1}{(\zeta_1)^2 + \cdots (\zeta_n)^2 + 1}, \cdots \frac{2\zeta_n}{(\zeta_1)^2 + \cdots (\zeta_n)^2 + 1} ,\frac{(\zeta_1)^2 + \cdots (\zeta_n)^2 - 1}{(\zeta_1)^2 + \cdots (\zeta_n)^2 + 1}\right)$$

I must be making a stupid computational error. Let's just look at the first coordinate. We get with $\zeta_1 = \frac{x_1}{1-x_{n+1}}$ that

$$\frac{2\zeta_1}{(\zeta_1)^2 + \cdots (\zeta_n)^2 + 1} = \frac{2\frac{x_1}{1-x_{n+1}}}{(\frac{x_1}{1-x_{n+1}})^2 + \cdots (\frac{x_n}{1-x_{n+1}})^2 + 1}$$

which after simplifying the bottom and cancelling gives

$$\frac{2x_1(1-x_{n+1})}{x_1^2 + \cdots x_n^2 + (1-x_{n+1})^2}$$

How does this simplify to $x_1$? I don't see it. So where's the mistake in computation?

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You do realize that your map is many-to-one and is not stereographic projection, right? – Will Jagy Sep 14 '12 at 3:11
Thanks. Your comment made me realize my error. This is supposed to be from $S^n \rightarrow R^{n+1}$, and the cancellation comes on the bottom since $|x| = 1$. – Euler....IS_ALIVE Sep 14 '12 at 3:13
see en.wikipedia.org/wiki/Stereographic_projection for low values of $n.$ – Will Jagy Sep 14 '12 at 3:17
No it's ok I got it now. Thanks though! – Euler....IS_ALIVE Sep 14 '12 at 3:19