Pentation Notation - How does it work? When going through with learning Grahams number, I got stuck at 
$$3↑↑↑3$$
Working it through, we have
$$3↑3=3^3$$
$$3↑↑3=3^{3^3}=3↑(3↑3)$$
As such, it would appear to me that
$$3↑↑↑3=3^{3^{3^3}}=3↑(3↑(3↑3))=3↑(3↑↑)$$
Which is incorrect; the correct answer being
$$3↑↑↑3=3↑↑(3↑↑3)$$
What I'm wanting to know is where the error in the way I've worked it through, and how working $3↑↑(3↑↑3)$ through backwards to $3↑↑↑3$ would look?
 A: You wish to understand Graham's number through these arrows? If so, I'd suggest stepping back down to multiplication and building the way up.
Note that
$$a\times b=\underbrace{a+(a+(\dots+a))}_b$$
For example,
$$3\times3=3+(3+3)=3+6=9$$
And then exponentiation,
$$a^b=a\uparrow b=\underbrace{a\times(a\times(\dots\times a))}_b$$
For example,
$$3\uparrow3=3\times(3\times3)=3\times9=27$$
Now tetration,
$$a\uparrow\uparrow b=\underbrace{a\uparrow(a\uparrow(\dots\uparrow a))}_b$$
For example,
$$3\uparrow\uparrow 3=3\uparrow(3\uparrow3)=3\uparrow27=7625597484987$$
And beyond...
$$a\uparrow\uparrow\uparrow b=\underbrace{a\uparrow\uparrow(a\uparrow\uparrow(\dots\uparrow\uparrow a))}_b$$
$$3\uparrow\uparrow\uparrow3=3\uparrow\uparrow(3\uparrow\uparrow3)=3\uparrow\uparrow7625597484987=\underbrace{3\uparrow(3\uparrow(\dots\uparrow3))}_{7625597484987}=3^{3^{3^{3^{\dots}}}}$$
A: $$ 3 \uparrow \uparrow n = 3 \uparrow (3 \uparrow \uparrow (n-1))$$
$$ 3 \uparrow \uparrow \uparrow n = 3 \uparrow \uparrow (3 \uparrow \uparrow \uparrow (n-1))$$
$$ 3 \uparrow \uparrow \uparrow \uparrow n = 3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow \uparrow (n-1)) ....$$
By definition 
$$ 3 \uparrow^{n} 0 =1\;\;\;\; 3 \uparrow 1 = 3 $$
One can prove by induction:
$$ 3 \uparrow^{n} 1 =3$$
$$ 3 \uparrow^{n} 2 = 3 \uparrow^{(n-1)} 3$$
$$ 3 \uparrow \uparrow 3 = 3 \uparrow {3^3} = 3^{3^3} = 3^{27} = 7625597484987$$
Therefore 
$$ 3 \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow (3 \uparrow \uparrow \uparrow 2)= 3 \uparrow \uparrow (3 \uparrow \uparrow 3) = 3 \uparrow \uparrow 7625597484987 $$
$$ 3 \uparrow \uparrow \uparrow 3 = 3^{3^{3^{3...}}}\;\;\; \text{tower height} = 3 \uparrow \uparrow 3 = 7625597484987$$
$$ 3 \uparrow \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow 3) $$
$$ ^{^{^{...3}3}3}3\;\;\; \text{tower height} = 3 \uparrow \uparrow 7625597484987 $$
