Find all $n \in \mathbb{N}^{*}$ such that $n^{n}+1$ is a prime number less than $10^{18}$.
We note that for $n^{n}+1$ to be less than $10^{18}$, it is necessary that $n10^{16} *(3 / 2)^{16}>10^{16} * 2^{8}>10^{16} * 100=10^{18}) . n=1$ works. If $n>1$ has an odd divisor $d>1, n^{n / d}+1 \mid\left(n^{n / d}\right)^{n}-(-1)^{d}=n^{n}+1$ and $11,2^{(} n k / d\right)+1$ divides $\left.\left(2^{(} n k / d\right)\right)^{d}+1=n^{n}+1$, and is strictly between 1 and $n^{n}+1$. Therefore, $\left.n=2^{(} 2^{p}\right)$. It remains to check the cases $n=2, n=4$, which work.