How do I write Grahams number I found that  Graham's number is :

So, can we say that it is equal to $3^x$ with $x$ is a power tower of 63 3's?
 A: Knuth's up-arrow notation works as follows :
$a\uparrow b=a^b$
$a\uparrow \uparrow b=a\uparrow a\uparrow ...\uparrow a \uparrow a$ with $b$ $a's$
$a\uparrow \uparrow \uparrow b=a\uparrow \uparrow  a \uparrow \uparrow ... \uparrow \uparrow a \uparrow \uparrow a$ with $b$ $a's$ 
and so on.
Note that the calculation is done from right , so $3\uparrow \uparrow 3=3\uparrow (3\uparrow 3)$, not $(3\uparrow 3)\uparrow 3$.
Now, Graham's number is defined as follows.
Notice the sequence
$G_0=4$ , $G_{n+1}=3\uparrow^{G(n)}3 $ for all $n\ge 0$
Then Graham's number is $G_{64}$.
Already $G_1=3\uparrow^4 3=3\uparrow \uparrow \uparrow\uparrow 3$ is so large that its magnitude cannot be comprehended. $G_2$ already has $G_1$ up-arrows,
$G_3$ has $G_2$ up-arrows and so on.
Just to imagine how big $G_1$ already is : First imagine the number
$$N:=3 \uparrow \uparrow 3 \uparrow \uparrow 3=3\uparrow \uparrow 3^{27}$$
This is a power tower of $3's$ with height $3^{27}$.
Now how to get $G_1$ :
Step $1$ : $3$
Step $2$ : $M_2 :$ a power tower with $3$ $3's$
Step $3$ : $M_3 :$ a power tower with $M_2$ $3's$
Step $4$ : $M_4 :$ a power tower with $M_3$ $3's$
and so on.
At step $N$, you reach $G_1$. It is already hard to describe how this number can be constructed. It is absolutely hopeless trying to comprend its size. Now you should get a feeling how insane big Graham's number is.
A: I wrote an article here describing the progression. Here is the progression the way I understand it:
$$ 3\uparrow3 = 3^3 = 27 $$  
$$ 3\uparrow\uparrow3 = 3\uparrow3\uparrow3 = 3^{3^3} = 7,625,597,484,987  $$
$$ 3\uparrow\uparrow\uparrow3 = 3\uparrow\uparrow3\uparrow\uparrow3 = 3\uparrow\uparrow(7,625,597,484,987) $$
Just to give you an idea of how big this is:
$$ 
\left.\begin{aligned}
3^{3^{3^{3^{⋰^{3}}}}} 
\end{aligned}\right\} 
\text{ height = 7,625,597,484,987}
$$
Now we can define G1:
$$ 3\uparrow\uparrow\uparrow\uparrow3 = 3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3 = 3\uparrow\uparrow\uparrow(3\uparrow\uparrow(7,625,597,484,987)) $$
Now we can use the result in G1 to define G2:
$$ G2 = \underbrace{3\uparrow\uparrow ... \uparrow\uparrow 3}_{G1 \text{ up arrows}} $$
$$ G3 = \underbrace{3\uparrow\uparrow ... \uparrow\uparrow 3}_{G2 \text{ up arrows}} $$
$$ G4 = \underbrace{3\uparrow\uparrow ... \uparrow\uparrow 3}_{G3 \text{ up arrows}} $$
$$ ...$$
This keeps going until you get to:
$$ G64 = \underbrace{3\uparrow\uparrow ... \uparrow\uparrow 3}_{G63 \text{ up arrows}} $$
And that's Graham's number. It's a HUGE number.
