$\DeclareMathOperator{\sgn}{sgn}$
The modulating signal in AM is
$$s(t) = C + a(t)\text,$$
where $a(t)$ is the (audio) amplitude, and $C$ is a constant so that $s(t) \ge 0 \;\forall t$. (Otherwise, your audio amplitude would just frequently "switch" the wave's sign, not really modulate the envelope.) That means, $C > - \min_t(s(t))$.
Therefore, the passband (around frequency $f$) signal is
$$r(t) = s(t) \sin(2\pi f t) = \left(C + a(t)\right) \sin(2\pi f t)\text.$$
As said, $C+a(t)$ is never negative; therefore the sign of $r(t)$ is
\begin{align}
\sgn(r(t)) &=\sgn(s(t) \sin(2\pi f t))\\
&=\sgn(s(t))\sgn(\sin(2\pi ft))\\
&=+1\cdot\sgn(\sin(2\pi ft))\\
&=\sgn(\sin(2\pi ft))
\end{align}
That tells us one way to write the absolute value of $r(t)$, because for any $x$:
\begin{align}
|x| &= x\cdot \sgn(x)\\
\implies\\
|r(t)|&=r(t)\cdot \sgn(r(t))\\
&= r(t)\cdot \underbrace{\sgn(\sin(2\pi ft))}_{:=w(t)}
\end{align}
This introduces us to a new function $w(t)$, which is always positive one when the carrier wave is, and negative when the carrier is negative. A square wave with the same frequency as the carrier! Thus, knowing that the square wave has a Fourier series representation $w_f(t)= \frac{4}{\pi} \sum\limits_{k\in\{1,3,5,\ldots\}} \frac{\sin\left(2\pi k ft\right)}k $
\begin{align}
|x| &= r(t)\cdot w(t) \\
&= r(t) \cdot \frac{4}{\pi} \sum\limits_{k\in\{1,3,5,\ldots\}} \frac{\sin\left(2\pi k ft\right)}k\\
&= \frac{4}{\pi} s(t) \sin(2\pi f t) \sum\limits_{k\in\{1,3,5,\ldots\}} \frac{\sin\left(2\pi k ft\right)}k\\
&= \frac4\pi s(t){\left[
\frac11\underbrace{\sin(2\pi f t)\sin(2\pi \,1f\, t)}_
{=\frac12(\cos(2\pi (f-f)t)-\cos(2\pi(f+f)t))}
+ \frac13\underbrace{\sin(2\pi f t) \sin(2\pi \,3f\, t)}_
{=\frac12(\cos(2\pi (f-3f)t)-\cos(2\pi(f+3f)t))}
+ \frac15\underbrace{\sin(2\pi f t) \sin(2\pi \,5f\, t)}_
{=\frac12(\cos(2\pi (f-5f)t)-\cos(2\pi(f+5f)t))} + \ldots
\right]}\\
&= \frac4\pi s(t)\left[{
{\frac12(1-\cos(2\pi2ft))}
+{\frac13\frac12(\cos(2\pi2ft)-\cos(2\pi4ft))}
+{\frac15\frac12(\cos(2\pi4ft)-\cos(2\pi6ft))} + \ldots}
\right]\\
&=\frac2\pi s(t)\left[{
{1-\cos(2\pi2ft)}
+\frac{\cos(2\pi2ft)-\cos(2\pi4ft)}3
+\frac{\cos(2\pi4ft)-\cos(2\pi6ft)}5 + \ldots}
\right]\\
&= \frac2\pi s(t) - \frac2\pi s(t)\left[{\frac23\cos(2\pi2ft)+\frac25{\cos(2\pi4ft)}+\frac27{\cos(2\pi6ft)} + \ldots}
\right]\\
\end{align}
That looks familiar: the first term in the square bracket, $\sin(2\pi f t)\sin(2\pi \,1f\, t)$ is simply mixing the carrier with the carrier, the second is mixing the carrier with a sine thrice its frequency and so on.
Thus, the first term yields, due to $\sin(a)\sin(b) =\frac12(\cos(a-b)-\cos(a+b))$, the baseband signal.
The rest of the terms mix up the passband to $2f$, $4f$, $6f$ and so on and are suppressed with a low-pass filter.
