For what values of $n$ is $10^8 < n! < 10^{12}$? This question is from the introduction of a handbook on combinatorics. So far, the material covered contains:


*

*counting problems solved by using trees, Pascal's triangle,...

*definition of $n$ factorial

*definitions of combinations, variations (based on allowing repitition and if order matters).


No formulas, except for the formula for $n$ factorial have been seen. 
The question then asks to find all values of $n$ such that 
$$10^8 < n! < 10^{12}$$
and the given answers states that this holds for $n = 12,13,14$ (only a numerical answer is given, no reasoning). 
I tried solving this by taking the logarithm with base 10 from all sides, but this did not really help (seemed to brute force to me). 
question: is there some 'nice' way to prove this answer?
 A: For the lower bound, observe that $$10^7=\underbrace{10\cdot 10\cdot ....\cdot 10}_{7\text{ times}}>10\cdot9\cdot8\cdot7\cdot6\cdot4\cdot3=9!\implies 10^8>10\cdot9!=10!$$
Thus $n>10$. Once noticed that $12!$ works (which isn’t very hard) observe the following for the upper bound 

if $a\in(10^8, 10^{12})$, then $10^4\cdot a\not\in (10^8, 10^{12})$

In particular $a\cdot 13\cdot 14\cdot 15\cdot 16\not\in (10^8, 10^{12})$ since $a\cdot 13\cdot 14\cdot 15\cdot 16>a\cdot 10^4$.
Hence, since $12!$ satisfies the inequality, $12!\cdot 13\cdot 14\cdot 15\cdot 16=16!$ doesn’t.
A: There's a nice way (called "Stirling's approximation") to get a narrower estimate, but I think that the goal here was to have you think about things. For instance, you know that $n > 8$, because at $n = 8$, the left hand side is 
$$
10 
\cdot 10 
\cdot 10 
\cdot 10 
\cdot 10 
\cdot 10 
\cdot 10 
\cdot 10 
$$
while the middle is 
$$
1 
\cdot 2
\cdot 3 
\cdot 4 
\cdot 5 
\cdot 6 
\cdot 7 
\cdot 8 
$$
Since each factor of the second is smaller than the corresponding factor of the first, you clearly need a larger $n$. 
On the other hand, $21!$ has $12$ factors that are at least $10$, namely $10, 11, 12, \ldots, 21$ (as well as a bunch of smaller factors). So $21!$ is definitely bigger than $10^{12}$. At this point, you know that $n$ is between $8$ and $21$, and trying a few numbers in between narrows things down pretty fast. 
But along the way, I expect the author hopes that you've learned something about estimating factorials by not-very-subtle methods. (The proof I know of Stirling's formula, for instance, involves integrals.)
A: By taking logs of both sides,
$$8\ln{(10)}\lt\ln{(n!)}\lt12\ln{(10)}$$
Then applying Stirling's approximation we have
$$8\ln{(10)}\lt n\ln{(n)}-n+\frac12\ln{(2\pi n)}\lt12\ln{(10)}$$
$$18.4...\lt n\ln{(n)}-n+\frac12\ln{(2\pi n)}\lt27.6...$$
Letting $f(x)=x\ln{(x)}-x+\frac12\ln{(2\pi x)}$ we have that
$$f(11)=17.4...$$
$$f(12)=19.9...$$
$$f(13)=22.5...$$
$$f(14)=25.1...$$
$$f(15)=27.8...$$
So the values of $n$ for which this is valid are $12,13$ and $14$.
