If $n\geq 2$ is prime and $a\geq 1$ is an integer, then $(a+1)^n-a^n$ leaves a remainder of $1$ modulo $p$.
If $n$ is even or a multiple of $3$, then it follows ad-hoc that again $(a+1)^n-a^n$ is never divisible by $n$.
By simulations, the same should be true for all $n$. How would one prove this?
Prove that $(a+1)^n-a^n$ is never divisible by $n$.
Assume, for the sake of contradiction, that there exists $n$ and $a$ such that $n$ divides $(a+1)^n-a^n$. Choose $n$ so that the smallest prime factor $p$ of $n$ is minimal (the OP has noted that $p\ge5$), and write $n=mp$. Then we have $(a+1)^{mp} \equiv a^{mp}\pmod p$, or equivalently $(a+1)^{m} \equiv a^{m}\pmod p$ by Fermat's little theorem, or equivalently again $(a+1)^{\gcd(m,p-1)} \equiv a^{\gcd(m,p-1)}\pmod p$. This implies that $\gcd(m,p-1)>1$; however, $m$ is not divisible by any primes smaller than $p$ by assumption, which contradicts $\gcd(m,p-1)>1$ since all primes dividing $\gcd(m,p-1)$ are at most $p-1$.
