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Repeated integration by parts gives the Gamma integralThe logarithm of its integrand is $n\log t-t$, whose unique maximum is at $t=n$. Write $t=n+\sqrt{n}s$. For fixed $s$ and large $n$, a Taylor expansion givesThe Big O notation here is for a fixed $s$, or uniformly for $s$ in a fixed bounded interval. It does not claim a uniform approximation over the entire growing integration domain. Consequently the peak, divided by its height, approaches $e^{-s^2/2}$.

The exact substitution givesReplacing the integral by the Gaussian integral ${\int }_{\mathbb{R}}{e}^{-s^2/2}ds=\sqrt{2\pi }$ produces $S_n$. Thus the estimate is peak height $n^n{e}^{-n}$ times width $\sqrt{n}$ times the area of a standard Gaussian peak. This is Laplace's method in a concrete form; controlling the integral's tails justifies the replacement. The next terms, including their products after exponentiation, produce the Stirling correction by Gaussian moments.

Watson's lemma applies to Laplace-type endpoint integrals such as ${\int }_0^b{e}^{-\lambda t}f(t)dt$ as $\lambda \to +\infty$, when $f$ has a local algebraic expansion $f(t)\sim {\sum }_j{c}_j{t}^{p+j-1}$ with $p>0$ and suitable integrability or exponential-growth control away from the endpoint. Each term contributes $c_j\Gamma (p+j){\lambda }^{-p-j}$. The exponential selects a neighborhood of $t=0$.

Laplace method instead treats ${\int }_a^{b}g(t)e^{\lambda \varphi (t)}dt$ by locating the dominant maxima of a real phase $\varphi$. A smooth nondegenerate interior maximum produces a neighborhood governed by a Gaussian function of width ${\lambda }^{-1/2}$; endpoint or degenerate maxima require their corresponding local scaling. Contributions away from a strict dominant maximum must be smaller. These methods overlap after suitable changes of variable, but their usual starting forms emphasize an endpoint amplitude expansion and an exponential maximum, respectively.

For the Gamma function, put $t=xu$. ThenThe phase has its unique maximum at $u=1$, with value $-1$ and second derivative $-1$. Set $u=1+s/\sqrt{x}$. Near this maximum,Combining the amplitude and exponential expansions in Laplace method givesLocalization near $u=1$ justifies extending the limits for the Gaussian integral; the distant endpoint is not being expanded uniformly by this local series. Odd terms integrate to zero. The moments of the Gaussian integral satisfy $I_1=I_0$, $I_2=3I_0$, $I_3=15I_0$, so the relative correction isThe two terms of the Stirling formula, obtained by the Stirling correction by Gaussian moments, are therefore

Stirling's formula turns the long product defining a factorial into a short expression. For positive integers $n$, putThen $n!\sim S_n$ as $n\to \infty$. This is asymptotic equivalence: $S_n/n!\to 1$, so the relative error tends to zero. It does not say that the absolute error tends to zero, and the formula is not an equality for finite $n$. The Gamma function extends this approximation beyond integer inputs. ^[ref]

For $n=10$,The leading estimate is about $0.830\%$ low. Multiplying by the first correction factor giveswhich is about $0.003176\%$ low. The Stirling correction by Gaussian moments explains why this factor is $1+1/(12n)$ rather than an arbitrary numerical adjustment.

The usefulness is in ratios as well as in enormous numbers. Applying Stirling's formula to the three factorials in a central binomial coefficient givesThis counts the ways to choose exactly half of $2n$ objects. Equivalently, for $2n$ independent fair coin tosses, the binomial distribution assigns probability $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\sim 1/\sqrt{\pi n}$ to exactly $n$ heads. An exponentially large count becomes a simple probability scale after cancellation.

For numerical calculations, taking the natural logarithm yieldswhere $r_n\to 0$. Computing a logarithm of a large factorial avoids representing the factorial itself. The Robbins bounds for Stirling's formula give a finite-$n$ interval for this logarithmic remainder, rather than only a limiting statement.

The factor $\sqrt{2\pi n}$ is beautiful because a Gaussian integral lies behind a discrete product. The Gaussian peak behind Stirling's formula displays this connection through the Gamma integral and Laplace's method. A rigorous proof must also control the tails of the integral; the local expansion explains the geometry but is not by itself a complete proof.

James Stirling published Methodus differentialis in 1730, a treatise on summation and interpolation of infinite series. This setting connects the formula with a practical calculation problem: taking logarithms turns a long product into a sum, whose growth can be approximated. ^[ref]