I'm about to prove that Graham's number is a number that is so big there is no way of intuitively grasping the magnitude of it, but in order to do this we must start with the very small and work our way up.
We will start with the number 9 which can represented as 3+3+3. This isn't so bad when you only have three 3's to write down, but what happens if I pick a number that isn't practical to be represented by +3's, like 729, I would need the sum of 243 3's, so now let's use multiplication. 729 can now be represented as 3*3*3*3*3*3 now I can represent larger numbers than before, but I run into the same problem. I can pick a number that isn't practical to be represented by the repetition of the same operator of multiplication.
7,625,597,484,987, for example, is the multiplication of 27 3's. That isn't good way to represent this number so now let's use powers to represent this number which is $3^{3^3}$, another way to represent this is $3 \uparrow 3 \uparrow 3$ where each single arrow is "to the power of operator".
Mathematicians realized when dealing with numbers that aren't practical to write down in one operator, a new operator is required that is more powerful the previous one. Arrow notation is a way of going from one operator to another with ease. $\uparrow \uparrow$ is the next operator up from $\uparrow$, just as $\uparrow$ is the next operator up from multiplication, and just as multiplication is one operator up from addition. So increasing the number of consecutive arrows will increase the ability to work with larger and larger numbers.
So we saw how the size of numbers that were created when using $\uparrow$. Now let's look at the $\uparrow \uparrow$. $3\uparrow \uparrow3\uparrow \uparrow3$ = $3\uparrow \uparrow(3\uparrow3\uparrow3)$=$3\uparrow \uparrow7,625,597,484,987$. Which means this is $\uparrow3$ is used on 3, 7,625,597,484,987 times!
${{{{{{{{{{3}^3}^3}^3}^3}^3}^3}^3}^3}^\cdot\Rightarrow}$ a stack of threes 7,625,597,484,987 high!
To calculate this value start with the top 3 and move downward.
$3^3 = 27$
$3^{27}=7,625,597,484,987$
$3^{7,625,597,484,987}$ is 3,638,334,640,024 digits long.
The fourth 3 is very large, but googolplex is 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 digits long. So still somewhat in our comprehension, but the fifth 3 is way out of our comprehension.
3 is raised to the power of a number that is 3,638,334,640,024 digits long.
Googolplex is 10 raised to the power of a number that is 101 digits long.
So we have used only 5 of the 3's on the 7,625,597,484,987 stack and it already becomes difficult to imagine, but this process is repeated over 7.6 trillion times. Doing this process 5 times took us from 3 to a number that makes googolplex look tiny. So the answer to the stack of 7,625,597,484,987 3's is a stupidly big number. This number is so big that if you memorized all the digits of this number your head would turn into a black hole. The maximum amount of entropy that can be stored in your head carries less information than the information of this stack of 3's.
[ http://m.youtube.com/watch?v=XTeJ64KD5cg ]
$$
\begin{array}{c|l|c|r}
\text{operator representation} & \text{value} & \text{number of previous operator equivalent} \\
\hline
3*3*3 & 27 &\text{9 (+3)'s or 3+3+3+3+3+3+3+3+3}\\
3\uparrow3\uparrow3& 7,625,597,484,987 &\text{27 (x3)'s or $3^{27}$} \\
3\uparrow\uparrow3\uparrow\uparrow3 & \text {stupidly big} & \text{7,625,597,484,987 ($\uparrow$ 3)'s}\\
3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3 & G_1 & \text {stupidly big ($\uparrow\uparrow3$)'s} \\
\end{array}
$$
If you look at the table above you will notice that $3\uparrow\uparrow\uparrow3\uparrow\uparrow\uparrow3=G_1$.
Why is $G_1$ important? Well I'll get to that in a minute. But first I have to explain that each change from one operation to the next is unimaginably bigger than previous change so you would think that adding a few arrows would quickly get us to Graham's number, but it doesn't. It doesn't even come close.
So what if I were to put a subscript on the arrows to indicate how many arrows I have. So $3\uparrow_{1000000}3$ is one million arrows, so I am one million operations above multiplication and 999998 operations above "stupid big" and the value of each operation is unimaginably bigger than the previous one. So, surely I must have hit Graham's number by now. The answer is no.
Remember $G_1$? Well, I'm going to write this, $3\uparrow_{G_1}3$.
So let's see if I can break this down.
"stupidly big" made googolplex look tiny after 5 of 7,625,597,484,987 iterations. Applying $\uparrow\uparrow3$ to 3 "stupidly big" times gives me $G_1$ and $G_1$ is now going to be the number of times the operator is increased. Where each operation is unimaginably bigger in comparison to the previous one. So how well do I stand against Graham's number at this point? Not even close.
Finally,
$G_2$=$3\uparrow_{G_1}3$
$G_3$=$3\uparrow_{G_2}3$
$G_4$=$3\uparrow_{G_3}3$
$\vdots$
$\vdots$
$G_{64}$=$3\uparrow_{G_{63}}3=\text{Graham's number}$
Best Answer
While it is true that one is much larger than the other in that we have $G_{\operatorname{TREE}(3)}-\operatorname{TREE}(3)$ being large, a linear scaling such as this is not very useful for comparing large numbers.
A simpler analogy would be $10^{1,000,000,000,001}$ and $10^{1,000,000,000,000}$. Although we have their difference being as big as $9\times10^{1,000,000,000,000}$, it probably does not seem like one is "significantly" bigger than the other.
One way to clarify what is meant here is by considering how the numbers are constructed, rather than the numbers themselves. When viewed from this perspective, it is clear to see that $10^{1,000,000,000,001}$ and $10^{1,000,000,000,000}$ are "made" the same way.
On the other hand, something such as $10^{10^{10^{10}}}$ is significantly larger than $10^{1,000,000,000,000}$ because it uses repeated exponentiation, which is far larger than just exponentiation.
In the same way, one could argue that
$$^{1,000,000,000,001}10=10^{10^{10^{.^{.^.}}}}\bigg\}1,000,000,000,001\text{ powers of }10$$
is not significantly larger than
$$^{1,000,000,000,000}10=10^{10^{10^{.^{.^.}}}}\bigg\}1,000,000,000,000\text{ powers of }10$$
Going back to our first example, one number was simply $10$ times larger than the other. In general, after a certain point, multiplying by $10$ is not significant. After a certain point, exponentiating one more time is not significant either. After a certain point, $G_n$ is not significantly larger than $n$.
To be more precise about how far one has to go for something to be considered insignificantly larger, we used in our examples:
This is, of course, very informal. Another way we could try to formulate this is with the fast growing hierarchy, as Peter mentioned, which can be made more formally.