I have the math formula:
$$\prod_{i = 0}^{v-1-1} \left (n-1-i \right) $$
which is in the solution reduced to:
$$\prod_{i = 0}^{v-1} \left (n-1-i \right) * \frac{1}{n-1-(v-1)} $$
What I don´t understand, is why is i replaced by (v-1) and which rule of reducing a Capital Pi is this, I can´t find a describtion why and how this is possible.
For a better understanding, here is the full solution:
The proof is in german, but that´s not necessary, I get anything except of why I can reduce the -1 and replace it with the i.
You have $$ \prod_{i = 0}^{v-1-1} \left (n-1-i \right) $$ Multiply that by $1$ written as a fancy fraction, giving $$ \frac{n-1-(v-1)}{n-1-(v-1)}\prod_{i = 0}^{v-1-1} \left (n-1-i \right) $$ and factor the numerator out as
$$ \frac1{n-1-(v-1)}\underbrace{(n-1-(v-1))\prod_{i = 0}^{v-1-1} \left (n-1-i \right)}_{\textstyle=\prod_{i = 0}^{v-1} \left (n-1-i \right)}$$
In other words, by changing the upper limit from $v-2$ to $v-1$ you're taking one more factor into the product; if you explicitly multiply by the inverse of that you get your original product up to $v-2$ back.