What is conditions for primality of numbers of form $N=10^n+1$?
If n is odd then we may write:
$$10^{2k+1}+1=(10+1)(10^{2k}-10^{2k-1}+ . . )$$
Therefore N can not be prime and we can conclude that the necessary condition for primality of N is even form of n.Now consider number $N=10^{2(2t+1)}+1$, where n is even, we have:
$$N=10^{2(2k+1)}+1=100^{2k+1}+1=(100+1)(100^{2k}+100^{2k-1}+ . . .)$$
That is in this case N is divisible by 101 and is not primes, or generally N in form of $N=10^{2^t(2k+1)}$ can not be prime.Hence we find the the sufficient condition is that the power of 10 must be of form $n=2^t$.So it seems $N=10^{2^n}+1$ is always prime.
Now consider numbers of form :
$$N=10^{2^{t_1}}+10^{2^{t_2}}+ . . . + 1$$
Some conditions of primality of this form are:
*Number of term must not be divisible by 9,; for N would be divisible by 9 and 3 otherwise.
*Number of terms must always be odd. For example number: $N=10^{16}+10^8+10^4+10^2+1=10^4(100^{3}+1)(100^2-100+1)+10^2(100^3+1)(10062-100+1)+1=101k+1$.
Are these conclusion correct? are there more conditions?
Your conclusion that n must be a power of $2$ is good. The next line saying that $N=10^{2^n}+1$ is always a prime is an incorrect step, which you haven't shown. In fact, it is false: $10^4+1 = 10001 = 73\cdot 137$.
More than three hundred years ago Fermat had a similar idea. He suggested $2^{2^n}+1$. Check out Fermat numbers / primes:
https://en.wikipedia.org/wiki/Fermat_number