Law of total probability - aplication

Question

I am trying to understand thesolution of the following example:

Each cuatomer who enters John's clothing store will purchase a suit with probability p. If the number of customers entering the store is a poisson random variable with mean $\mu$, what is the probability that John sells k suits?

Solution:

$P(X=k) = \sum_{n=0}^\infty P(X=k|N=n) P(N=n) = \sum_{n=0}^\infty\binom{n}{k} p^k (1-p)^{n-k}e^{-\mu}\frac{\mu^n}{n!} $

So far i could understand. Bue the rest of the solution i can not understand. Ould someone be able to break it down for me ?

$ = e^{-\mu}\frac{(\mu p)^{n}}{k!} \sum_{n=0}^\infty \frac{(\mu(1-p))^{n-k}}{(n-k)!} = e^{- \mu p} \frac{(\mu p)^k}{k!},k>0 $

Answer 1

Note that $P(X=k\mid N=n)=0$ if $n<k$ so that we can start the summation with $n=k$ instead of $n=0$.

Let's abbreviate $q:=1-p$

$$\sum_{n=k}^{\infty}\binom{n}kp^kq^{n-k}e^{-\mu}\frac{\mu^n}{n!}=e^{-\mu}\sum_{n=k}^{\infty}\binom{n}k(\mu p)^k(\mu q)^{n-k}\frac1{n!}=e^{-\mu}\sum_{n=k}^{\infty}\frac{(\mu p)^k(\mu q)^{n-k}}{k!(n-k)!}=$$$$e^{-\mu}\frac{(\mu p)^k}{k!}\sum_{n=k}^{\infty}\frac{(\mu q)^{n-k}}{(n-k)!}=e^{-\mu}\frac{(\mu p)^k}{k!}\sum_{n=0}^{\infty}\frac{(\mu q)^{n}}{n!}=e^{-\mu}\frac{(\mu p)^k}{k!}e^{\mu q}=e^{-\mu p}\frac{(\mu p)^k}{k!}$$

Answer 2

$$\sum_{n=0}^{\infty} \binom{n}{k}p^k (1-p)^{n-k} e^{- \mu} \frac{\mu^n}{n!} = e^{-\mu} p^k \sum_{n=0}^{\infty} \frac{n!}{k!(n-k)!} (1-p)^{n-k}\frac{\mu^n}{n!} = e^{-\mu}\frac{(p \mu)^k}{k!} \sum_{n=0}^{\infty} \frac{(\mu (1-p))^{n-k}}{(n-k)!} = e^{-\mu}\frac{(p \mu)^k}{k!} e^{\mu (1-p)} = e^{-\mu p} \frac{(p \mu)^k}{k!}$$