@Anon-e-moose

Take all the matter in the universe, grind it up to make ink and it still wouldn't be enough to write out Graham's Number

That's because there are less particles in the observable universe than g_{64}.

yet, I ends in a '7'.

That's provable with modular arithmetic. First note that g_{64} is a power of 3. The basic heuristic is,

3^{0}≡1 (mod 10)

3^{1}≡3 (mod 10)

3^{2}≡9 (mod 10)

3^{3}≡7 (mod 10)

3^{4}≡1 (mod 10)

3^{5}≡3 (mod 10)

...

And the pattern repeats; 1, 3, 9, 7, 1, 3, 9, 7, etc. It can be seen that 3^{n}≡1 (mod 10) if n is congruent to 0 mod 4, 3^{n}≡3 (mod 10) if n is congruent to 1 mod 4, 3^{n}≡9 (mod 10) if n is congruent to 2 mod 4, and lastly 3^{n}≡7 (mod 10) if n is congruent to 3 mod 4.

So now we can determine the last digit of 3^{n} by looking at the remainder of n divided by 4.

This type of argument can be made upwards through the power-tower defining g_{64} and will determine the last digit of the number.

What is the last decimal digit of Pi, Andy?
...oh, and while we're waiting, here is proof that Dividing By Zero equals Infinity:

I know you're joking but division by zero is undefined in the complex numbers and π has no last digit in base 10. Although it would be nice if Andy would try to do an impossible task like that and not report back to the internet until he is finished.