Maybe Prime!

Let: $N = 10^{10^{10^n}} + 10^{10^n} + 10^n - 1.$ Write $n = 2^m k$ with $m$ a nonnegative integer and $k$ a positive odd integer. For any nonnegative integer $j$ , $10^{2^m j} \equiv (-1)^j \pmod{10^{2^m} + 1}.$ Since $10^n \geq n \geq 2^m \geq m+1$ ,

$10^n$ is divisible by $2^n$ and hence by $2^{m+1}$ , and similarly $10^{10^n}$ is divisible by $2^{10^n}$ and hence by $2^{m+1}$ . It follows that $N \equiv 1 + 1 + (-1) + (-1) \equiv 0 \pmod{10^{2^m} + 1}.$ Since $N \geq 10^{10^n} > 10^n + 1 \geq 10^{2^m} + 1$ , it follows that $N$ is composite.

If we expand the terms just for imagination then then the whole expression will looks like

$1000......000100.....00001...0000-1=1000......000100.....00001...0009$ .We are not interested about the number of zeroes.This big number contains only three 1's and a single 9.So, the sum of the digits is $1+1+1+9=12 \equiv 0 \pmod{3}$ .So, the whole expression is divisible by $3$ .

Therefore for every $n$ the above expression is not a prime.