Primes

Since

$16^{16}+1≈1.8\cdot 10^{19}>10^{19}$

We get the bound for $a$

$1\leq a \leq 15$

If $a$ is an odd number, $a^a+1$ is even and it is surely composite, exept for the case $a=1$ that gives the prime $2$

Now suppose $a$ was not a power of $2$ and let $a=2^b\cdot c$ , where $2\nmid c$ , $c\neq 1$ , so

$\begin{aligned} a^a+1&=a^{2^b\cdot c}+1\\ &=\left(a^{2^b}\right)^c+1\\ &=\left(a^{2^b}+1\right)\left[\left(a^{2^b}\right)^{c-1}-\left(a^{2^b}\right)^{c-2}+\ldots+1\right] \end{aligned}$

Which is not prime, so it is necessary but not sufficient that $a$ is a power of $2$ . We have only $3$ powers of $2$ to check

$2^2+1=5\phantom{ccccccc}\text{Prime}$

$4^4+1=257\phantom{ccccccc}\text{Prime}$

$8^8+1=2^{24}+1=\left(2^8\right)^3+1=256^3+1=(256+1)(256^2-256+1)\phantom{ccccccc}\text{Not prime}$

Hence the product of all primes of the form $a^a+1$ less than $10^{19}$ is

$2\cdot 5\cdot 257=\boxed{2570}$