Phonon Distribution Factor in Phonon Emission Rate

The rate at which a phonon with wavevector $$\vec{q}$$ is absorbed is given by $$\frac{1}\tau \propto n(\hbar \omega(\vec{q}))$$ This is pretty obvious to me. The more phonons there are the more often the absorption.

Meanwhile, the rate at which a phonon with wavevector $$\vec{q}$$ is emitted can be written as $$\frac{1}\tau \propto [1+n(\hbar \omega(\vec{q}))]$$

I would like to know the reasoning behind this. If able, I want a physical interpretation to this.

What I thought of so far:

1. $$[1+n(\hbar \omega(\vec{q}))]$$ is equal to $$-n(-\hbar \omega(\vec{q}))$$, but I am not sure how to interpret this.
2. Since it is an emission process, right when the emission happens the number of existing phonons with wavevector $$\vec{q}$$ increases by one.

If we start in a state with $$n$$ phonons and calculate the probability that one phonon is emitted, we typically have a matrix element like $$\langle n+1|b^\dagger|n\rangle=\sqrt{n+1}\langle n+1|n+1\rangle=\sqrt{n+1}.$$
On the other hand, if a phonon is absorbed, the matrix element will have a factor $$\langle n|b|n+1\rangle=\sqrt{n}\langle n|n\rangle=\sqrt{n}.$$
Averaging over Boltzmann distribution transforms $$n$$ into the phonon distribution function.
Note also that in the classical regime, where the number if phonons is very big, one can neglect $$1$$.