How to calculate the coefficient of $$$\prod\limits_{i=0}^{p-1}(x+i) \bmod p$$$ ($$$p$$$ is a prime)? Thank you for reply.
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It is also the first kind of Stirling number.
which coefficient r u interested in?
In fact the answer is very beautiful, but sadly no one mentioned it.. It equals to $$$x^p-x$$$. I don't know why, can anyone explain it?
From Fermat's theory we know that $$$x^p \equiv x\pmod p$$$ holds for every integer. So when modulo $$$p$$$, $$$F(x)=x^p-x$$$ must be a multiple of $$$x,x+1,…,x+(p-1)$$$ at the same time. Pay attention to the fact that $$$F$$$ is of degree $$$p$$$, and the given equation is proved.
Now, I'm not a math genius or anything.
However, I do know that when multiplying a list of X consecutive numbers, the result is always divisible by X. Therefore, the answer to your question is 0, because the result is always divisible by P, even if P is a prime.
I want to calculate the coefficient for every $$$x^i$$$, and the answer is $$$x^p-x$$$, that there is only two nonempty positions.