Among some permutation polynomials I've been studying, $f(x)=(x+1)^n-x^n$ is one of the polynomials that I cannot grasp.

Question: For a givenodd prime$p$, how can we findevery positive integer$n$ such that $f(x)=(x+1)^n-x^n$ is a permutation polynomial mod $p$ ?

The answer seems $n=(p-1)m+2\ \ (m=0,1,2,\cdots)$, but I'm facing difficulty in proving that. This question has been asked previously on math.SE without receiving any answers.

The followings are what I've got.

$f(0)\equiv 1.$

$f(p-1)\equiv -(-1)^n\Rightarrow \text{$n$ has to be even}\Rightarrow f(p-1)\equiv p-1$.

For $n=(p-1)m+r$, $f(x)\equiv (x+1)^r-x^r$ because $a^{p-1}\equiv 1$ for $a$ which is coprime to $p$.

$f\left(\frac{p-1}{2}\right)\equiv 0$.

$f\left(\frac{p-1}{2}+a\right)+f\left(\frac{p-1}{2}-a\right)\equiv 0$ for any $a$.

I would like to know any relevant references as well.