Sum of independent Poisson distributions and binomial theorem

Question

I have two independent Poisson variables $X_1$ and $X_2$ for which I have calculated the sum in convolution. I obtained:

$$ e ^{\mu_1 \mu_2}\sum_{k=O}\frac{\mu_1^k}{k!}\frac{\mu_z^{z-k_1}}{z-k_1!}$$

My professor has then used the binomial formula $ (a+b)^n = \sum \frac {n!}{k!(n-k)!}a^k b^{n-k}$ to come up with the solution that:

$$ \sum_{k=O}\frac{\mu_1^k}{k!}\frac{\mu_z^{z-k_1}}{z-k_1!} = \frac{(\mu_1 + \mu_2)^z}{z!}$$

I do not understand this last step, could someone break it down for me ?

Answer 1

The LHS is $\frac{1}{z!}\sum_k\binom{z}{k}\mu_1^k\mu_2^{z-k}$, which by the binomial theorem is $\frac{1}{z!}(\mu_1+\mu_2)^z$.