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Here are collected some results related with the operator $a$ which commutation with its adjoint $\ud{a}$ reads:
$$\tag{1}[a,\ud{a}]=1$$
The basic rules are:
$$ \begin{align*} a\ket{n}&=\sqrt{n}\ket{n-1}\,,&\bra{n}\,&a=\bra{n+1}\sqrt{n+1}\,,\\ \ud{a}\ket{n}&=\sqrt{n+1}\ket{n+1}\,,&\bra{n}\,&\ud{a}=\bra{n-1}\sqrt{n}\,, \end{align*} $$
of which a general expression can be drawn:
$$a^{\dagger i}a^j a^{\dagger k}\ket{n}={(n+k)!\over(n+k-j)!}\sqrt{(n+i+k-j)!\over n!}\ket{n+i+k-j}\,.$$
Some particular cases:
$$ \begin{align*} \kern-1cm{(\mathrm{for}~i\le n+j)}\kern1cm a^i{\ud{a}}^j\ket{n}&={(n+j)!\over\sqrt{n!}\sqrt{(n+j-i)!}}\ket{n+j-i}\,,\\ \kern-1cm{(\mathrm{for}~i\le n)}\kern1cm a^{\dagger j}a^i\ket{n}&={\sqrt{n!}\sqrt{(n+j-i)!}\over(n-i)!}\ket{n+j-i}\,. \end{align*} $$
A general result, based on Wick theorem:
$$a^n\ud{a}^m=\sum_{k=0}^{\min(n,m)}k!\binom{m}{k}\binom{n}{k}\ud{a}^{m-k}a^{n-k}$$
Some useful particular cases:
$$a\ud{a}^n=\ud{a}^na+n\ud{a}^{n-1}$$ $$a^n\ud{a}=\ud{a}a^n+na^{n-1}$$
A compilation of useful results derived from (1)
We define $\hat n\equiv\ud{a}a$.
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Special cases below
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