To be clear, the other end digits to consider are 3,5,9, but the pattern proof for 7 yields same result for all numbers, where a final digit of 1 ultimately reduces to (1) by *(3) and /(2^2)... Ending in (9)*(3)=7+1, 8 divided by (2^3)=1. Ending in 5 proves 3.. (5)*(3)=5+1, 6/2 yields either 8 or 3 depending on second(tens) digit even/odd. 8/(2^3)=1, so only need to prove 3. (3)*(3)=9+1, 0/2 = 5,, somebody else can elequently show this ([even][6])/(2) pattern for digits {3,5} ultimately yields a trailing (8) from final digits being ([odd][6])/(2) and thus easily reduces to (1). Adam Slattery -helo Fri-13-dec-2019