The answer is (B). Let us consider the case in which $n$ is odd. Then any power of it is odd, and the following number is even. The only even prime number is $2$, and it is possible to obtain it only in the case $n=1$. Let us now consider the case in which $n$ is even. In this case $n+1$ is odd, and it is therefore possible to carry out the factorization $n^{n+1}+1=(n+1) \cdot Q(n)$. If $n$ is an even number different from $0$, $n+1$ is a number greater than $2$ and $Q(n)$ a number greater than $1$, hence $n^{n+1}+1$ is not a prime number.