Trying to understand Wick's theorem, I took most of my knowledge from the corresponding Wikipedia article. The statement is that given the definition of normal ordering of operators $A,B,C,\ldots$ any product of those operators can be expressed as \begin{align} ABCDEF\ldots=\;&:ABCDEF\ldots: \\\\ &+\sum_{\mathrm{singles}} :A^\bullet B^\bullet CDEF\ldots: \\\\ &+\sum_{\mathrm{doubles}} :A^\bullet B^{\bullet\bullet} C^{\bullet\bullet}D^\bullet EF\ldots: \\\\ &+\ldots \end{align}
where $:ABCDEF\ldots:$ is the normal ordering of the operators and $A^\bullet B^\bullet\equiv AB-:AB:$ is their contraction.
Problem: Consider the quantum harmonic oscillator with ladder operators $a,a^\dagger$ and $N=a^\dagger a$ with commutation relations $[a,a^\dagger]=1$ and $[N,a]=-a$ and $[N,a^\dagger]=a^\dagger$. I am interested in changing the order of terms like $N^na$. Therefore I define the normal ordering as $:Na:=aN$ and hence $N^\bullet a^\bullet=[N,a]=-a$.
Now apply Wick's theorem to the term $N^na$ and we find \begin{align} N^na =& :N^na:+:N^{n-1}N^\bullet a^\bullet:+:N^{n-2}N^\bullet Na^\bullet:+\ldots \\\\ =&:N^na:+n:N^{n-1}N^\bullet a^\bullet: \\\\ =&aN^n-naN^{n-1} \end{align} where I made use of the fact that only contractions involving $a$ survive so that we only get terms with single contractions.
Apparently this is not the correct result, as a simple example with $n=2$ shows \begin{align} N^2a=&aN^2+[N^2,a] \\\\ &=aN^2+N[N,a]+[N,a]N\\\\ &=aN^2-Na-aN \\\\ &=aN^2-2aN-[N,a] \\\\ &=aN^2-2aN+a \end{align} where I used the fact that $[A^2,B]=A[A,B]+[A,B]A$. This result differs from the one obtained using Wick's theorem: $N^2a=aN^2-2aN$.
I am sure that the mistake lies in my understanding how to apply the theorem. Can someone point out what I did wrong?
