See YouTube or wikipedia for the defination of Graham's number.

A Googol is defined as $10^{100}$.

A Googolplex is defined as $10^{\text{Googol}}$.

A Googolplexian is defined as $10^{\text{Googolplex}}$.

Intuitively, it seems to me that Graham's number is larger (maybe because of it's complex definition).

Can anybody prove this?

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    $\begingroup$ Graham's number is much much bigger. As in, take 3^3^3^3^... for the rest of your life, and you aren't even remotely close in any sense. $\endgroup$ – user233746 Dec 18 '16 at 13:21
  • $\begingroup$ Better yet, have everyone alive from now until the universe ends, help you take 3^3^3^3^... as fast as possible. You still won't be able to reach Grahams number. $\endgroup$ – user233746 Dec 18 '16 at 13:27
  • $\begingroup$ Even iterating the number of $3$'s in the power tower $3\uparrow3 \uparrow\cdots \uparrow 3$ a googol times (starting with $3^{3^3}$) will not get even close to Graham's number $\endgroup$ – Peter Dec 19 '16 at 11:40

Googolplex can be bounded from above like a tower of exponents, so $$10^{10^{100}} < (3^3)^{10^{100}} = 3^{3\times 10^{100}} <3^{10^{101}} < 3^{(3^3)^{101}} = 3^{3^{303}} < 3^{3^{3^{3^3}}}$$ In the last step, we have used the fact that $303$ is much,much smaller than $3^{27}$. Now, take the Googolplexian.We can thus, easily check that, $$10^{10^{10^{100}}} < (3^3)^{10^{10^{100}}} < 3^{3^{3^{3^{3^3}}}}$$ So, Googolplexian is much smaller than a tower of exponents of $3$'s of length $6$, or in other words Googolplexian is less than $3\uparrow \uparrow6$.(using Knuth's up-arrow notation.)

Now, compare this with just the first layer of Graham's number,i.e., $3\uparrow \uparrow \uparrow \uparrow 3$. Hope it helps.

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  • $\begingroup$ After the first equality sign It should be $3^{3\times 10^{100}}$ instead of $3^{3\times 10^{10^{100}}}$. $\endgroup$ – timon92 Dec 18 '16 at 9:33
  • $\begingroup$ @timon92 Thank you for the correction. $\endgroup$ – user371838 Dec 18 '16 at 9:36
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    $\begingroup$ The inability to even conceive what $3\uparrow\uparrow\uparrow\uparrow3$ is should be enough. Or even drop an arrow. $\endgroup$ – Simply Beautiful Art Dec 18 '16 at 13:43

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