How big are factorials?



The other day, I found myself wondering how big 52! (52 factorial) is, and that led me to ponder how these could be estimated without a calculator or a computer.

It turns out there’s some fairly interesting math behind being able to estimate the size (number of digits) of a factorial reasonably accurately. This post will start by stating how to do the estimate, and if you’re curious you can read on for the math background.

Without further ado, the approximation is:

\[\text{number of digits in n!}\approx n\log_{10}\left(\frac{n}{e}\right)+2\]

As an example, let’s use my original question, by estimating this for 52!

Well, 52 divided by e is... 20-ish? And \log_{10}(20) is about 1.3 [1]; therefore our estimate comes out to:

\[\text{number of digits in 52!}\approx 52\cdot 1.3 +2 \approx69\]

The real answer is 68, so this is very close! In estimates like this - when you’re dealing with enormous numbers - being off by a couple of digits usually isn't a big deal.

The Gamma function

The Gamma function for real n>0 is defined [2] as:

\[\Gamma(n)=\int_{0}^{\infty}x^{n-1}e^{-x}dx\]

This integral does not have an analytic expression in the general case, but it does have a very useful property that we can take advantage of. Let’s see what \Gamma(n+1) is:

\[\Gamma(n+1)=\int_{0}^{\infty}x^{n}e^{-x}dx\]

And now use integration by parts with:

\[u=x^n\qquad v=-e^{-x}\]

Then:

\[du=nx^{n-1} dx\qquad dv=e^{-x}dx\]

So:

\[\begin{aligned} \Gamma(n+1)&=\int_{0}^{\infty}x^{n}e^{-x}dx\\ &=\left. -x^n e^{-x}\right|_{0}^{\infty}-\int_{0}^{\infty}-e^{-x}n x^{n-1}dx\\ &=n\int_{0}^{\infty}x^{n-1}e^{-x}dx \end{aligned}\]

But notice that the last integral is just \Gamma(n); therefore, we’ve shown that:

\[\Gamma(n+1)=n\Gamma(n)\]

Let’s also calculate \Gamma(1) - it’s a special case that has an analytical solution:

\[\Gamma(1)=\int_{0}^{\infty}e^{-x}dx=\left. -e^{-x}\right|_{0}^{\infty}=1\]

This helps establish an induction argument:

\[\begin{aligned} \Gamma(2)=1\cdot\Gamma(1)&=1!\\ \Gamma(3)=2\cdot\Gamma(2)&=2!\\ \Gamma(4)=3\cdot\Gamma(3)&=3!\\ \dots\\ \Gamma(n+1)=n\cdot\Gamma(n)&=n! \end{aligned}\]

In other words - the Gamma function is an interpolation of the factorial over all positive reals. Here’s a plot of the Gamma function over a small range; note that the y axis is log-scale because of the function’s fast growth:

Gamma function plot

Stirling’s approximation

You may have encountered Stirling’s approximation before:

\[n!\approx\sqrt{2\pi n}\cdot\left(\frac{n}{e} \right)^n\]

It’s a great approximation that works reasonably well even for small n. This section is a brief overview of how Stirling’s formula is derived from the Gamma function.

Taking:

\[n!=\Gamma(n+1)=\int_{0}^{\infty}x^n e^{-x}dx\]

We’ll start by massaging the integrand a bit:

\[n!=\int_{0}^{\infty}x^n e^{-x}dx=\int_{0}^{\infty}e^{n \ln x} e^{-x}dx\]

And making a change of variables x=ny, which means that dx=ndy:

\[\begin{aligned} n!&=\int_{0}^{\infty}ne^{n\ln(ny) - ny}dy=n\int_{0}^{\infty}e^{n(\ln n+\ln y-y)}dy\\ &=ne^{n\ln n}\int_{0}^{\infty}e^{n(\ln y-y)}dy \end{aligned}\]

These steps make the integral amenable to applying Laplace’s method, which allows us to approximate definite integrals of the form:

\[\int_{a}^{b}e^{nf(x)}dx\]

Where f(x) is a twice-differentiable function and n some large number. By Laplace’s method, such integrals can be approximated by:

\[\int_{a}^{b}e^{nf(x)}dx\approx\sqrt{\frac{2\pi}{n|f''(x_0)|}}e^{n f(x_0)}\]

Where x_0 is the global maximum of f(x).

Let’s see how to apply this method [3] to the latest equation we have for n! (renaming the dummy integration variable back to x):

\[n!=ne^{n\ln n}\int_{0}^{\infty}e^{n(\ln x-x)}dx\]

In our case, f(x)=\ln x - x. It’s easy to show that this function is twice differentiable and has a global maximum at x_0=1. Moreover:

\[\begin{aligned} f(x_0)&=-1\\ f''(x_0)&=-1\\ \end{aligned}\]

Substituting these into the proper places in Laplace’s approximation, we get:

\[\begin{aligned} n! &\approx ne^{n\ln n}\sqrt{\frac{2\pi}{n}}e^{-n}\\ &\approx \sqrt{2\pi n}\cdot e^{n(\ln n - 1)}\\ &\approx \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n\quad\blacksquare \end{aligned}\]

Number of digits from Stirling’s approximation

We can calculate the number of digits in n! by taking the base-10 logarithm of Stirling’s formula:

\[\begin{aligned} \text{number of digits in n!}&\approx \log_{10} \left(\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n\right)\\ &\approx \log_{10}\left(\sqrt{2\pi n}\right)+ \log_{10}\left(\frac{n}{e}\right)^n\\ &\approx \log_{10}\left(\sqrt{2\pi n}\right)+ n\cdot \log_{10}\left(\frac{n}{e}\right) \end{aligned}\]

Note that the first term is not multiplied by n itself; therefore, as n grows, it will become less and less noticeable. That said, it still adds a couple of digits - so you should take it into account if you want a more accurate approximation [4]

[1]Mental tricks for calculating \log_{10} is a different topic, but it really helps to remember that \log_{10}2=0.3, \log_{10}3=0.5, and from here using the various logarithm laws to estimate multiples.
[2]As a matter of fact, the Gamma function is defined for complex numbers, but for our purpose talking about the reals is sufficient.
[3]As much as I’d like to dive into why this approximation works, the rabbit hole in this post is already deep enough!
[4]It adds up to 2 extra digits as long as n is less than 1600 or so, and may add more than 2 after that, though no more than 3 until n is 160000. It’s not clear why anyone would like to estimate the number of digits of 1600! (about 4450, in case you were wondering), let alone 160000!

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