7. When $n=1$, $n^{n}+1=2$ satisfies the condition. When $n>1$, let $n=2^{k} q$, where $q$ is an odd number. If $q>1$, as shown in the previous problem, $n^{n}+1$ is not a prime number, so $n=2^{k}$, where $k$ is a positive integer. At this point,
$$n^{n}+1=2^{k \cdot 2^{k}}+1=\left(2^{2^{k}}\right)^{k}+1$$

Further analysis shows that there exists a non-negative integer $m$ such that $k=2^{m}$, hence
$$n^{n}+1=2^{2^{2^{m}+m}}+1$$

When $m \geqslant 2$, $2^{m}+m \geqslant 6$, so $2^{2^{m}+m} \geqslant 2^{6}$, thus
$$\begin{aligned} n^{n}+1 & \geqslant 2^{2^{6}}+1=2^{64}+1 \\ & =16 \times(1024)^{6}+1 \\ & >16 \times\left(10^{3}\right)^{6}+1 \\ & >10^{19} \end{aligned}$$

Therefore, by $n^{n}+1 \leqslant 10^{19}$, we know $m \leqslant 1$. By setting $m=0,1$, we find $n^{n}+1=5,257$, both of which are prime numbers.

In summary, the required prime numbers are 2, 5, and 257.