Simple formula for the $n$-ary version of $(x,y) \mapsto \frac{x+y}{1-xy}$ Let $x * y = \frac{x + y}{1 - xy}$. I want a single formula for $x_1 * x_2 * \ldots * x_n$, for all natural $n$.
In order to generate plausible candidates, let's see what happens at small values of $n$:


*

*$x_1 * x_2 * x_3 = \dfrac{x_1 + x_2 + x_3 - x_1x_2x_3}{1 - x_1x_2 - x_1x_3 - x_2x_3}$

*$x_1 * x_2 * x_3 * x_4 = \dfrac{x_1 + x_2 + x_3 + x_4 - x_1x_2x_3 - x_1x_2x_4 - x_1x_3x_4 - x_2x_3x_4}{1 - x_1x_2 - x_1x_3 - x_1x_4 - x_2x_3 - x_2x_4 - x_3x_4 + x_1x_2x_3x_4}$


Then I conjecture the following:
$x_1 * x_2 * \ldots * x_n = \dfrac
{\sum_{I \in T_1} \prod_{i \in I} x_i - \sum_{I \in T_3} \prod_{i \in I} x_i}
{\sum_{I \in T_0} \prod_{i \in I} x_i - \sum_{I \in T_2} \prod_{i \in I} x_i}$
Where $T_k = \{\, I \in \wp(\{1, 2 \ldots n\}) : k \equiv |I| \pmod 4 \, \}$
Is there a nice proof of the above not involving transcendental functions?
 A: Expanding on my comment...
Suppose all the $x_j$ are real.  We proceed by induction.  We write $\frac{\Im}{\Re}(t)$ to represent the ratio of the imaginary to real parts of the complex number $t$.  Then $$\begin{align}
x_1 &= \frac{\Im}{\Re}( (1+\mathrm{i} x_1) ) = x_1 \text{ and }\\
x_1 \ast x_2 &= \frac{\Im}{\Re}( (1+\mathrm{i} x_1)(1+\mathrm{i} x_2) ) = \frac{x_1 + x_2}{1 - x_1 x_2},
\end{align}$$ establishing the result for the smallest allowed $n$.  (Actually, I've assumed this is what you want for $n=1$, since you didn't specify.)  Suppose now $n >2$ and $\ast_{j=1}^{n-1} x_j = \frac{\Im}{\Re}\left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)$.  Then 
$$\begin{align}
    \frac{\Im}{\Re}\left( \prod_{j=1}^{n} (1 + \mathrm{i}x_j) \right)
    &= \frac{\Im \left( (1 + \mathrm{i}x_n) \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)}{\Re\left( (1 + \mathrm{i}x_n) \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)} \\
    &= \frac{\Im \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) + x_n \Re \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) }{\Re \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) - x_n \Im \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)} \\
    &= \frac{ \frac{\Im}{\Re} \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) + x_n }{1 - x_n \frac{\Im}{\Re} \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right)} \\
    &= \frac{\Im}{\Re} \left( \prod_{j=1}^{n-1} (1 + \mathrm{i}x_j) \right) \ast x_n \\
    &= \left( \ast_{j=1}^{n-1} x_j \right) \ast x_n \\
    &= \ast_{j=1}^{n} x_j.
\end{align}$$  (It's traditional to write this last display in reverse order, but I think the manipulations are easier to follow in this order.)  The step yielding $\dots {} - x_n \Im \dots$ in the denominator may be less than obvious:  we want the negative imaginary part here because (in the previous line) when it is multiplied by $\mathrm{i} x_n$, it must yield a positive real part.  (Trying to stumble across this manipulation in the traditional order is why I think the order above is easier to follow.)  This completes our induction.
Having done that, it should be no great challenge to show that $\frac{\Im}{\Re}\left( \prod_{j=1}^{n} (1 + \mathrm{i}x_j) \right)$ splits the even and odd degree terms into the denominator and numerator, respectively, and also has the signs you want for your $\pmod{4}$ representation.
