About the transformation of a formula (of probability products) I am reading a paper and getting stuck with some involved formula transformations. Here comes the settings of the problem corresponding to the formula:  
Initially, we have a fixed parameter vector $θ=(θ_1,θ_2,θ_3), ∑θ_i =1$
We have the following generative story:  


*

*Draw $n\sim Poisson(\lambda)$

*With respect to $n$, we have a $Multinomial(n,\theta)$

*Call $C$ a set of all vectors $D=(d_1,d_2,d_3)$ that satisfy:   $D \sim Multinomial(n,\theta)$ and $d_1\times d_2 \times d_3 \neq 0  $  


The problem then accounts to the calculation of this summation:
$$S=∑_{(n=3)}^∞(\space p(n|\lambda )×∑_{(D∈C)} p(D |n,θ)\space )$$
The authors introduced this formula :
$$S=e^{-\lambda}∏_{i=1}^3(e^{λ×θ_i} -1)$$
I have been working on the formulas to understand how did the authors present the above formula. However, with the modest skill on summation, I haven't made it out yet.
I would highly appreciate any clue or instruction on how to come up with the above shortened formula(you don't need to perform the transformation explicitly) 
Please feel free to ask me if you need further clarifications 
Thank you!
 A: The sum, which initially appears to be a quadruple sum, extends over all $n\ge 3$, $d_1 \ge 1$, $d_2\ge 1$, and $d_3\ge 1$, where $n=d_1+d_2+d_3$.  Therefore it's really just a triple sum over all integral tuples $(d_1,d_2,d_3)$ with coordinates extending from $1$ upwards.
Moreover, from the Poisson formula
$$p(n\mid\lambda) = e^{-\lambda} \frac{\lambda^n}{n!}$$
and the Multinomial formula
$$p(D\mid n,\theta) = \frac{n!}{d_1!d_2!d_3!} \theta_1^{d_1}\theta_2^{d_2}\theta_3^{d_3}$$
the cancellation of $n!$ in numerator and denominator and the identity $\lambda^n=\lambda^{d_1}\cdots\lambda^{d_3}$ allow us to simplify the product--which is the quantity being summed--as
$$p(n\mid\lambda)\,p(D\mid n,\theta)  = e^{-\lambda}\frac{\lambda^{d_1}\theta_1^{d_1}}{d_1!} \cdots \frac{\lambda^{d_3}\theta_3^{d_3}}{d_3!} = e^{-\lambda}\frac{(\lambda\theta_1)^{d_1}}{d_1!}\cdots \frac{(\lambda\theta_3)^{d_3}}{d_3!}.$$
Consequently, applying the series expansion $$e^x - 1 = \sum_{d=1}^\infty \frac{x^d}{d!}$$ to the cases $x=\lambda \theta_i$, we may write
$$\eqalign{
S &= \sum_{d_1=1,d_2=1,d_3=1}^\infty e^{-\lambda}\frac{(\lambda\theta_1)^{d_1}}{d_1!}\cdots \frac{(\lambda\theta_3)^{d_3}}{d_3!}\\&=e^{-\lambda}\sum_{d_1=1}^\infty\frac{(\lambda\theta_1)^{d_1}}{d_1!}\cdots\sum_{d_3=1}^\infty\frac{(\lambda\theta_3)^{d_3}}{d_3!}\\&=e^{-\lambda}\left(e^{\lambda\theta_1}-1\right)\cdots\left(e^{\lambda\theta_3}-1\right)\\&= e^{-\lambda}\prod_{i=1}^3 \left(e^{\lambda\theta_i}-1\right).
}$$
