Graham's number stable digits: An exact solution

In the decimal numeral system, we prove that the well-known Graham's number, $G := \! ^{n}3$ (i.e., $3^{3^{\cdot^{\cdot^{\cdot^{3}}}}}$ ($n$ times)), and any base $3$ tetration whose hyperexponent is larger than $n$ share the same $\operatorname{slog}_3(G) - 1$ rightmost digits (where $\operatorname{slog}$ indicates the integer super-logarithm). This is an exact result since the $\operatorname{slog}_3(G)$-th rightmost digit of $G$ differs from the $\operatorname{slog}_3(G)$-th rightmost digit of $^{n+1}3$. Furthermore, we show that the $\operatorname{slog}_3(^{n}3)$-th least significant digit of the difference between Graham's number and any base $3$ tetration whose integer hyperexponent exceeds $n$ is $4$.