= Solution
First include the mean-gain calculation preceding the numbered clauses. At each <electron-multiplying CCD> stage one electron contributes an expected $1+p$ electrons to the next stage. If $N_j$ is the charge after stage $j$, <conditional expectation> gives $\mathbb E[N_{j+1}\mid N_j]=(1+p)N_j$. Iteration from $n$ input electrons yields
$$
\boxed{g=(1+p)^r,\qquad \mathbb E[x_n]=ng.}
$$
For small $p$, $g\simeq e^{rp}$ if corrections of order $rp^2$ are negligible.
In the high-gain approximation the output from one input electron has an <exponential distribution> of scale $g$. Independent input electrons produce independent cascades, and their charges add. By <convolution of independent random variables>, assuming the formula for $n$ inputs,
$$
P_{n+1}(x)=\int_0^x P_n(y)P_1(x-y)\,dy
=\frac{e^{-x/g}}{g^{n+1}(n-1)!}\int_0^x y^{n-1}\,dy
=\frac{x^ne^{-x/g}}{g^{n+1}n!}.
$$
Together with the single-electron base case this proves
$$
\boxed{P_n(x)=\frac{x^{n-1}e^{-x/g}}{g^n(n-1)!},\qquad x\geq0,\ n\geq1.}
$$
It is a <gamma distribution> with shape $n$ and scale $g$, so it is normalized and has <expected value> $ng$ and <variance> $ng^2$. The formula is a continuous approximation to discrete output charge, not an exact integer-valued branching distribution; for $n=0$ there is a point mass at zero instead.
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