7:59 AM
@Mathphile $n^n-81$ can only be semiprime in the following cases :
$(1)$ : $n$ is odd and $$\frac{n^n-81}{2}$$ a prime number
$(2)$ : n is even and $n^{n/2}-9$ and $n^{n/2}+9$ are both prime.
I did not find an example yet in either case.
If we allow a negative semiprime, $n=2$ is however a solution
$(1)$ is not satisfied for $n\le 5\ 600$ , so the number must have more than $21\ 000$ digits in this case.