Deriving MGF for binomial distribution

Question

Using the definition of the binomial distribution, I obtain that: $$\Psi (t) = (pe^t+q)^n $$ I then compute $\Psi ' = npe^t(pe^t+q)^{n-1}$

I then evaluated this at $\Psi'(0)$ and got $\Psi'(0)=np(p+q)^{n-1}$

and so $E[X]$ which I know is $np$.

How do I then differentiate $\Psi'(t)$ again, the $n-1$ is throwing me off?

Answer 1

Put p+q=1( by property of binomial distribution), and you get E(X)=np