Let create a function $f(n,p)=(n+1)^p-n^p$. So we can be sure, that $f(n,p)$ can be expressed as $$f(n,p)=(2ap+1)(2bp+1)(2cp+1)\cdots$$ where $2kp+1$ - prime factors of $f(n,p)$and $a,b,c,\cdots$ - natural numbers (if $n>0$, $p$ - odd prime).
How can I prove it? How is it useful? Is there any formulas of quantity of primes $2pk+1$ from $1$ to $m$?
If I made some mistakes, sorry for my English.
On
Let $q$ be a prime divisor of $(n+1)^p - n^p$. Then obviously we have that $q \not \mid n$, as then $0 \equiv (n+1)^p - n^p \equiv 1 \pmod q$. So now let $n^{-1}$ be the multiplicative inverse of $n$ modulo $q$. Then we have:
$$(n^{-1}(n+1))^p \equiv 1 \pmod q$$
Now the order of $n^{-1}(n+1)$ modulo $q$ can't be $1$, as then $n^{-1}$ is the multiplicative inverse of $n+1$, which is impossible. Therefore the order of $n^{-1}(n+1)$ modulo $1$ is $p$, as $p$ is prime. Therefore we have that:
$$p \mid q-1 \implies q = tp + 1$$
As both $q$ and $p$ are odd we must have that $t=2k$ and so $q=2kp + 1$ and hence the proof.
Let $q$ be a prime dividing $(n+1)^p-n^p$. Of course we have $$(n+1)^p\equiv n^p\pmod q$$
Now, if $p$ does not divide $q-1$ then $a\mapsto a^p$ is an automorphism of the multiplicative group $\pmod q$; in particular it is injective.
It follows that we must have $p\,| \, q-1$.
Note too that $(n+1)$ and $n$ have opposite parities so $(n+1)^p-n^p$ is odd, and we are done.