Why do we do the transformation of the variable in this proof for Poisson distribution?

Question

I have 2 questions related to this question.

1. Why do we do the transformation for $i$ to $j$ in the form of $i = j+1$ in the 4th line of the equation? It seems to me here like it is either done so that we can get the equation back to original Poisson form with a different variable or to correct for the summation interval. I still cannot intuitively understand it.
2. In the 6th line, it is written as $\lambda\operatorname E((X+1)^{n-1}).$ But I feel like it should be $\lambda\operatorname E((Y+1)^{n-1})$ because did we not transform the variable in the 4th line of the equation

Answer 1

1. It's to get back the Poisson form. The second equality comes from noticing you don't need the $i=0$ term because it's zero. The fourth inequality comes from relabelling to get back the Poisson form in $j$ instead of $i$

2. Yes, but no, but it doesn't matter. The last equality goes from a sum over $j$ to the expectation of a Poisson random variable, and it would be true for any Poisson($\lambda$) random variable; it only depends on the distribution. That is the equality of $E[X^n]$ and $\lambda E[(X+1)^{n+1}]$ doesn't depend on whether $X$ is the same variable in the two expressions. Because it doesn't, you might award math points for using a different letter ($Y$) in the two expressions, even though it doesn't matter. However, the fourth equality holds exactly because $i$ and $j+1$ are always the same, so in fact the two $X$s are the same random variable even though we don't need that fact, so actually you might award math points for calling them both $X$.