Let X be a Poisson random variable, show that $E[X^n] = \lambda E[(X+1)^{n-1}]$ I am struggling to understand some of the steps of the proof of this problem:

Let X be a Poisson random variable, show that $E[X^n] = \lambda
 E[(X+1)^{n-1}]$.

The proof:
$$
\begin{align*}
E(X^n) &\stackrel{(1)}=
e^{-\lambda} \sum_{j=0}^\infty j^n \frac{\lambda^j}{j!} \\ &\stackrel{(2)}=
e^{-\lambda} \sum_{j=0}^\infty j^n \frac{\lambda^j}{j!} \\ &\stackrel{(3)}=
e^{-\lambda} \sum_{j=1}^\infty j^{n-1} \frac{\lambda^j}{(j-1)!} \\ &\stackrel{(4)}=
\lambda e^{-\lambda} \sum_{j=1}^\infty j^{n-1} \frac{\lambda^{j-1}}{(j-1)!} \\ &\stackrel{(5)}=
\lambda e^{-\lambda} \left( \sum_{i=0}^\infty (i+1)^{n-1} \frac{\lambda^i}{i!} \right) \\ &\stackrel{(6)}=
\lambda E((X+1)^{n-1}).
\end{align*}
$$
Assuming the lines are numbered 1-6, I will try to explain and show where I'm stuck (the bolded parts):

*

*This is just by definition, but I don't understand why it is necessarily going from 0 to infinity, and why does it look for $P(X=i), 0\leq i\leq \infty $ (hope my questions was clear)


*Since for $j=0$ it doesn't contribute to the sum, I can omit it and count from 1


*This line is not clear to me, why can I reduce the power and the denominator? doesn't make much sense to me


*Nothing special here


*We change the lower bound again to count from 0, why again I can simply add 1 to the power and denominator?


*Why is this conclusion correct?
Thanks in advance!
 A: $1$ - the expected value of a function $f(x)$ of a discrete random variable $X$ is $\sum\limits_x f(x)\, \mathbb P(X=x)$. Here $f(x)=x^n$. For a Poisson random variable with $\mathbb P(X=x) = e^{-\lambda}\dfrac{\lambda^x}{x!}$, the possible values of $X$ are the non-negative integers, so the sum is over the non-negative integers 
$3$ - By dividing both the numerator and the denominator by non-zero $j$ you can say $\dfrac{j^n}{j!} = \dfrac{j^{n-1}}{(j-1)!}$
$5$ - if you do the substitution $i=j-1$, then $j^{n-1}=(i+1)^{n-1}$ and $\lambda^{j-1}=\lambda^i$ and $(j-1)!=i!$ and $j=1 \implies i=0$
$6$ - this is the reverse of the first point: you have inside the brackets $\sum\limits_x g(x)\, \mathbb P(X=x)$ so this is the expectation of $g(x)=(x+1)^{n-1}$
A: *

*This is like for any set of numbers the sum of the numbers is taken when finding their mean. This comes from the definition of expected value.

*Sure.

*You're simply cancelling the $j$ from $j^n$ with the $j$ from $\frac{1}{j!}$. That is, $j^n = j^{n-1}j$ and $j! = j(j-1)!$.

*Okay.

*This is a basic change of index. The basic form for this is $\sum_{k=2}^{\infty}k = \sum_{j=1}^{\infty}(j+1)$.

*We substitute line 5 into the original form to get this result.

