How to calculate $e^{ad\hat{A}}\hat{B}$?

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For $\hat{a}$ and $\hat{a}^{\dagger}$ is annihilation and creation operators which, $[\hat{a},\hat{a}^{\dagger}]=1$.

Could you please show me the way to calculate,

$\left(e^{-i\omega\hat{a}^{\dagger}\hat{a}t}\right)^{\dagger}\hat{a}\left(e^{-i\omega\hat{a}^{\dagger}\hat{a}t}\right)=e^{ad(-i\omega\hat{a}^{\dagger}\hat{a}t)}\hat{a}=\hat{a}e^{i\omega t}$

P.S. This is a homework-exercise problem. I was given to solve the above relation plus other like, $e^{ad(-i\omega\hat{a}^{\dagger}\hat{a}t)}\hat{a}^{\dagger},e^{ad(-i\omega\hat{a}^{\dagger}\hat{a}t)}\sigma_{\pm},e^{ad(-i\omega\hat{a}^{\dagger}\hat{a}t)}\sigma_{z},...$. But I think I could do the rest if I know how to calculate one. Thank you.

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Hints:

(1) $\text{ad}\left(\hat{a}^\dagger\hat{a}\right)\,\hat{a}=-\hat{a}$.

(2) $\Big(\exp\big(-\text{i}\omega\hat{a}^\dagger\hat{a}t\big)\Big)^\dagger=\exp\big(\text{i}\omega\hat{a}^\dagger\hat{a} t\big)$, assuming that $\omega$ and $t$ are real numbers (which are the usual settings).

(3) $\exp\big(\text{i}\omega\hat{a}\hat{a}^\dagger t\big)\,\hat{a}=\hat{a}\,\exp\big(\text{i}\omega\hat{a}^\dagger\hat{a}t\big)$.

(4) $\exp\left(\hat{A}\right)\,\exp\left(-\hat{A}\right)=1$.