How big is a particular n!? Is there a way to estimate how big $n!$ is for a certain $n$?
For example, without using a calculator, what is the magnitude of $7!$ or $12!$ or $100!$?
 A: A different tack here, as you I suspect you are asking for a method that doesn't require any 'advanced' functions.
How large a positive integer number is, can be expressed well in terms of how many digits it takes to write the number. This is best represented as the integer part of the logarithm of the number, plus 1 (logarithm base 10, i.e. Log not Ln). For example LOG[55] = 1.74 (2 d.p.), it requires INT[1.74] + 1 = 2 digits to represent it. You can get a better feel for the size of a number by looking at the digits after the decimal point in the logarithm too.
Doing polynomial fits of LOG[n!] vs n gives these two useful approximate working regions.
n = 7 to 119 (cubic polynomial): LOG[n!] ~ -0.0000284659105*n^3 + 0.00943671908*n^2 + 0.962859106*n - 3.95423829
Maximum error factor in range of 2.93
n = 120 to 1000 (5th order polynomial): LOG[n!] ~ -4.68977059E-13*n^5 + 0.00000000164012606*n^4 - 0.000002402405*n^3 + 0.0021512217*n^2 + 1.6770136*n - 29.6620872
Maximum error factor in range of 1.94
You could do better with higher order polynomials, particularly in range 7 to 119, but these are good enough to tell you "how big" the factorials are.
Below 7 you should be able to do in your head :-)
A: Yes, Stirling's formula:
\begin{equation}
n! \; \sim \; \sqrt{2\pi n} \left(\frac{n}{e}\right)^n.
\end{equation}
A: Another answer in this thread suggests Stirling's approximation: $$n! \approx \sqrt{2\pi n}\biggl(\frac ne\biggr)^n$$
which I think is the right place to start for any answer to this question.  However, it may be useful to add that if you take the logarithm of this amount, you get the (approximate) number of digits you need to write down $n!$, which may be of more use:
$$\begin{align}
\log n! 
& \approx \color{darkblue}{n\log n} - n\color{maroon}{\log e} + \color{darkblue}{\frac12 \log n} + \color{darkgreen}{\frac12\log 2\pi}\\
& \approx \color{darkblue}{\biggl(n+\frac12\biggr)\log n} - \color{maroon}{0.43}\cdot n + \color{darkgreen}{0.798}
\end{align} $$
($\log$ here is the common, base-10 logarithm function.)
For small $n$ this gives the following values, which I have rounded off to the nearest integer:
$$\begin{array}{rr}
n & \text{length of $n!$} \\\hline
 1 &        0 \\
 2 &        1 \\
 3 &        1 \\
 4 &        2 \\
 5 &        2 \\
 6 &        3 \\
 7 &        4 \\
 8 &        5 \\
 9 &        6 \\
10 &        7 \\
11 &        8 \\
12 &        9 \\
13 &       10 \\
14 &       11 \\
15 &       13 \\
16 &       14 \\
17 &       15 \\
18 &       16 \\
19 &       18 \\
20 &       19 \\
\end{array}
$$
This is correct for all $n$ shown except $1$ and $5$, for which it is off by 1.
A: A cool alternative to Stirling's formula could be the more accurate formula derived by Srinivasa Ramanujan,
$$
\ln(n!) \sim n\ln(n) - n + \frac{1}{6}\ln(n(1+4n(1+2n))) + \frac{1}{2}\ln(\pi).
$$
