Computation of $e^{i \hbar \omega a^{\dagger} a} a e^{-i \hbar \omega a^{\dagger} a}$ I need to compute terms like :
$$e^{i \omega t a^{\dagger} a} a e^{-i  \omega t a^{\dagger} a}$$
Where $[a,a^{\dagger}]=1$ (they are the bosonic annihilation/creation operators).
I wonder if there is a simple formula for this. Indeed, when I try to compute the commutator:
$$[a,e^{i \omega t a^{\dagger} a}]. $$ 
I end up with something that doesn't look trivial.
For example:
$$[a^{\dagger} a, a] =a .$$
But:
$$[(a^{\dagger} a)^2, a] =2 a^{\dagger} a^2 $$
So I don't know how I could compute the general term (and if actually it is an easy thing to do...).
In summary: is there a simple expression for:
$$e^{i \omega t a^{\dagger} a} a e^{-i  \omega t a^{\dagger} a}$$
and if so, is there a trick to compute it?
 A: Hint: there is a general identity 
$$ \exp(\hat{X})\hat{Y}\exp(-\hat{X}) = \hat{Y} + \left[\hat{X},\hat{Y}\right] + \frac{1}{2!}\left[\hat{X},\left[\hat{X},\hat{Y}\right]\right] + \frac{1}{3!}\left[\hat{X},\left[\hat{X},\left[\hat{X},\hat{Y}\right]\right]\right] + ...\ ,$$
which I believe would be useful for your purposes.
A: I define the following operator:
\begin{equation}\tag{1}
\hat{A}(t) = U^{\dagger} a \, U = e^{i \omega t \hat{N}} a \, e^{- i \omega t \hat{N}},
\end{equation}
where $\hat{N} \equiv a^{\dagger} a$.  We have the following commutator (notice that there's a sign mistake in yours): 
\begin{equation}\tag{2}
[a, a^{\dagger}] = \mathbb{1}, \quad \Rightarrow \quad [\hat{N}, a] = - a.
\end{equation}
Then we have this:
\begin{align}
\frac{d \hat{A}}{dt} = i \omega \, [\hat{N}, \hat{A}] &= i \omega \, U^{\dagger}[\hat{N}, a] \, U \\[12pt]
&= - i \omega \, U^{\dagger} a \, U = - i \omega \hat{A}. \tag{3}
\end{align}
The solution to this differential equation is easy:
\begin{equation}\tag{4}
\hat{A}(t) = A(0) \, e^{- i \omega t} = a \, e^{- i \omega t}.
\end{equation}
