Which is bigger: $9^{9^{9^{9^{9^{9^{9^{9^{9^{9}}}}}}}}}$ or $9!!!!!!!!!$? In my classes I sometimes have a contest concerning who can write the largest number in ten symbols. It almost never comes up, but I'm torn between two "best" answers: a stack of ten 9's (exponents) or a 9 followed by nine factorial symbols. Both are undoubtedly huge, but I haven't been able to produce an argument that one is larger (surely they aren't equal). Any insight into which of these two numbers is bigger would be greatly appreciated.
 A: I find it quite sad that none of the answers has actually formally shown that 

$$9^{9^{9^{9^{9^{9^{9^{9^{9^9}}}}}}}} > 9!!!!!!!!!$$

The proces is a bit long, but it it worth it in my opinion. First, a simple calculation shows $9! < 9^8$. Then $9!! < (9^8)^{9^8} = 9^{8\cdot9^8}$. Then $$9!!!< \left(9^{8\cdot9^8}\right)^{9^{8\cdot9^8}} < \left(9^{9^9}\right)^{9^{8\cdot9^8}}=9^{9^{8\cdot9^8}\cdot9^9}=9^{9^{8\cdot9^8+9}}$$
$$9!!!! < \left(9^{9^{9^9}}\right)^{9^{9^{8\cdot9^8+9}}}=9^{9^{9^{8\cdot9^8+9}}\cdot 9^{9^9}}=9^{9^{9^{8\cdot9^8+9}+9^9}}<9^{9^{9^{8\cdot9^8+10}}}$$
Now, some notation. Let us use $f(x)$ for $9^x$ and $f^k(x)=f(f(... f(f(x))...))$. Let $x!^k$ denote $x!!!...!!!$ with $k$ factorials. 
Clearly, $f(x)+1<f(x+1)$. By induction, $f^t(x)+1<f^t(x+1)$ for any $t$. 
Now we proceed by induction to show that $9!^k < f^{k-1}(8\cdot9^8+k+6)$. The base case has been given above. Assume $4 \leq k \leq 9^8-7$. 
$$9!^{k+1} < \left(f^{k-1}(8\cdot9^8+k+6)\right)^{f^{k-1}(8\cdot9^8+k+6)}=
9^{f^{k-2}(8\cdot9^8+k+6)f^{k-1}(8\cdot9^8+k+6)}<9^{9^{f^{k-3}(9^9)+f^{k-2}(8\cdot9^8+k+6)}}<9^{9^{9\cdot f^{k-2}(8\cdot9^8+k+6)}}<9^{9^{9^{f^{k-3}(8\cdot9^8+k+6)+1}}}<9^{9^{9^{f^{k-3}(8\cdot9^8+k+7)}}}=f^k(8\cdot9^8+k+7)$$
This completes the induction hypothesis. Hence we have shown the asked.

Now for some fun: The largest number using ten symbols. We can also use shortened Knuth's up arrow notation, $9 \uparrow^{9 \uparrow^{9 \uparrow^{9} 9} 9} 9$ has ten symbols.
Conway chained arrow notation, $9 \rightarrow 9 \rightarrow 9 \rightarrow 9 \rightarrow 9^9$.
Rado's sigma function, $\Sigma(\Sigma(\Sigma(9)))$. 
A: $n!$ grows more rapidly than $9^n$ (it grows approximately with $n^n$), so eventually it wins out; in fact, a few moments with Wolfram Alpha suggests that for all $n\gt 21$, $n! \gt 9^n$.  This means that the best solution is mixed: after the first exponentiation (since $9^9$ is greater than $9!$), you're better off using factorials, giving the answer $9^9!!!!!!!!$.  (Also, note that we're all assuming that the correct parentheses here are implicit, since there's a big difference between $\left(9^9\right)^9$ and $9^{\left(9^9\right)}$.)
On the other hand, if you're allowed to use more-or-less stock mathematical notation, you may want to have a look at Knuth's arrow notation; $9\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow\uparrow9$ is bigger by far than the rest of these.  (And numbers even larger than this have come up in mathematics - have a look, for instance, at Graham's Number, which most conveniently uses the arrow notation in its definition.)
EDIT: In fact it's pretty straightforward to prove that the answer in the first paragraph, $9^9!!!!!!!!$, is the best that can be done (with these operations).  First, as long as $a$ is at least two symbols (for this group of symbols), then $a^9 \lt 9^a\lt a!$, so adding a factorial symbol is always going to be better than adding a $9$ onto your tower.  But what about structures like $9!^{9!}$?  Well, if both $a$ and $b$ are at least two symbols long, then $a^b\lt \mathrm{max}(a,b)!!$; assuming for the moment that $b$ is larger (because if $a$ were larger, then $b^a\gt a^b$), then $b! \approx b^b$, and while this isn't enough to ensure that it's greater than $a^b$ (consider the case where $a$ = $b$ = $9!$), it's close enough that we can be certain taking a second factorial will be larger - so those extra symbols you spent on exponentiation should retroactively have just gone into more exclamation points, and this is enough to guarantee that $9^9!!!!!!!!$ is the largest possible combination of these particular operations.
A: Wolfram Alpha shows $$9^{9^9}\approx 10^{10^{8.56}}$$ while $$(9!)! \approx 10^{10^{6.26}}$$  Adding another character adds another 10 to the bottom of the stack without changing the upper exponent much at all.  It will even do the full stack of 10.  The exponents have 8.5678 atop the tower of 10's, while the factorials have 6.26949.  So the exponents win.
Added:  In Douglas Hofstadter's May, 1982 column "On Number Numbness" he declares these essentially equal.  For numbers of this size, the first thing you should look at is how many times you have to take a log to make it reasonable, which is 9 for both of them.  Then look at the number on top of the stack, which is the only one that matters.  So it is like 9.85678 compared with 9.626949, which are very close.
A: A thorough study of this question is given in the article "Exponential vs. Factorial" by Velleman (American Mathematical Monthly Vol. 113, No. 8, Oct. 2006, pp. 689-704).  In the motivating example there are 5 characters instead of 10.  Of course, the conclusion is the same as in Steven Stadnicki's answer, that $9^9!!!!!!!!$ (or $9^9!!!$ in the case of 5 characters) is the largest possible.
The article also addresses your original question, proving that the exponential tower of $n$ $9$s is always larger than a $9$ with $n-1$ factorials applied (and many more general statements).
