We have $N_k={{100^k-1}\over {99}}$ and we look at $k>1$. If $k$ is odd, then $N_k={{10^k+1}\over {11}}\cdot {{10^k-1}\over 9}$ hence not prime. If $k$ is even then $k=2^a l$ with $a\ge 1, l\equiv 1(2)$ and $10^{2k}-1=10^{2^{a+1}l}-1=(10^{2^al}+1)\cdots (10^l+1)(10^l-1)$. In the factors with +1 only the amount of factors 2 decreases. Dividing by 99 we get ${{10^l+1}\over {11}}\cdot {{10^l-1}\over 9}$ in the end, hence it can only be prime if $l=1,a=1,k=2, N=101$ which is indeed prime.