How to approximate $49^{4}81^{5}$? I am doing this exercise for the GMAT test

Which of the following option is closest to $49^{4}81^{5}$?
A. $8^{18}$
B. $8^{19}$
C. $8^{20}$
D. $8^{21}$
E. $8^{22}$

My attempt:

$49^{4}81^{5} = (7^2)^4(9^2)^5 = 7^{8}9^{10} = (8-1)^{8}(8+1)^{10} \approx 8^{8}8^{10} = 8^{18}$

But I am not sure how $(8-1)^{8}(8+1)^{10}$ is closer to $8^{18}$ than to $8^{19}$.
Please shed me some lights. Thank you for your help!

Let's take it exactly for a few more steps:
$$
(8-1)^{8}(8+1)^{10} =(8-1)^8(8+1)^8(8+1)^2\\
=(8^2-1)^8\cdot9^2<8^{16}\cdot 9^2
$$
And $9^2$ is much closer to $8^2$ than to $8^3$.
