Don't forget that the last digit could also be $0$, but there are no such prime numbers (since anything ending in $0$ is divisible by $10$). Otherwise, yes, this is true:
If you have a string of final digits $a_n \cdots a_1$ with $a_1 \neq 0,2,4,5,6,8$, then the number $k = 10^{n-1} a_n + \ldots + a_1$ is not divisible by $2$ or $5$ so is prime to $10^n$. By a celebrated theorem of Dirichlet, the arithmetic progression $k, k + 10^n, k + 2 \cdot 10^n,\ldots$ then contains infinitely many prime numbers. Any one of these is congruent to $k$ modulo $10^n$, which is equivalent to saying that the sequence of the last $n$ digits is $a_n \cdots a_1$.