Proof of Wick's theorem I'm tackling proof of Wick's theorem. By induction.
Let us suppose we have already proved
$$ C_2 \cdots C_n = N(C_2 \cdots C_n + (\text{all possible contractions}) ) \quad (C_i=a\,\, \text{(annihilation) or }a^{\dagger}\text{(creation)} \,\, ) , $$
and multiply $C_1$ from the left.
Then we move $C_1$ inside $N()$.
Though the real computation manipulates all the terms inside $N()$, let us consider the term with no contractions.
If $C_1=a$,
\begin{alignat}{2}
C_1 N(C_2 \cdots C_n) &=&& a (a^{\dagger} \cdots a^{\dagger} a \cdots a) \\
&=&& N(:C_1 C_2:C_3 \cdots C_n \, +\, C_2:C_1 C_3: \cdots C_n\,+ \cdots\cdots\,+\, C_2\cdots C_{n-1} :C_1 C_n: ) \\
&&&+ (a^{\dagger} \cdots a^{\dagger} a \cdots a)a \\
&=&& N( :C_1 C_2:C_3 \cdots C_n\, +\, C_2:C_1 C_3: \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} :C_1 C_n:\,+\,C_1 \cdots C_n  )
\end{alignat}
there's no problem.
But if $C_1=a^{\dagger}$,
\begin{alignat}{2}
C_1 N(C_2 \cdots C_n) 
&=&& a^{\dagger} (a^{\dagger} \cdots a^{\dagger} a \cdots a) \\
&=&& N(:C_1 C_2:C_3 \cdots C_n \,+ \,C_2:C_1 C_3: \cdots C_n \,+\cdots\cdots\,+\, C_2\cdots C_{n-1} :C_1 C_n: ) \\
&&&+ (a^{\dagger} \cdots a^{\dagger} a \cdots a)a^{\dagger} 
\end{alignat}
then the last term of the last line isn't equal to $N( C_1 \cdots C_n)$. 
Why?
To get better insight, I calculated a example.
From
$$ a_2 a_3^{\dagger} = N(a_2 a_3^{\dagger} + :a_2 a_3^{\dagger}: ) , $$
(I added subscripts just for comprehensibility. So for instance $a_2^{\dagger}=a_3^{\dagger}$.)
I got
\begin{alignat}{2}
a_1^{\dagger} a_2 a_3^{\dagger} 
&=&& a_1^{\dagger} a_3^{\dagger}a_2 + a_1^{\dagger}:a_2 a_3^{\dagger}: \\
&=&& N(a_1^{\dagger} a_2 a_3^{\dagger} + a_2 : a_1^{\dagger} a_3^{\dagger}: + a_1^{\dagger}:a_2 a_3^{\dagger}: ) .
\end{alignat} 
But according to Wick's theorem, the relation
$$ a_1^{\dagger} a_2 a_3^{\dagger} 
= N(a_1^{\dagger} a_2 a_3^{\dagger} + : a_1^{\dagger} a_2: a_3^{\dagger} + a_2 : a_1^{\dagger} a_3^{\dagger}: + a_1^{\dagger}:a_2 a_3^{\dagger}: ) $$
should hold.
How does the term $: a_1^{\dagger} a_2: a_3^{\dagger}$ emerge?
Thanks.
 A: Let's consider the case in which there exist $h$ pairs of $d_i, d_i^{\dagger}$. 
$$d_1, \cdots , d_h ,d_1^{\dagger} \cdots d_h^{\dagger}$$
$$ [d_i, d_j^{\dagger}] = \delta_{ij}$$
In this formula and induction, subscripts denote the class of operators. In the example below($a_i$'s), they doesn't. 
The definition of contraction is
$$ <0|d_i d_j|0> =0,\, <0|d_i d_j^{\dagger}|0> = \delta_{ij}\,
<0|d_i^{\dagger} d_j|0> =0\, <0|d_i^{\dagger} d_j^{\dagger}|0> =0. $$
From
$$ C_2 \cdots C_n = N(C_2 \cdots C_n + (\text{all possible contractions}) ) \quad (C_i=d_j\,\, \text{(annihilation) or }d_j^{\dagger}\text{(creation)} \,\, ) , $$
If $C_1=d_i$, using
$$d_i d_j^{\dagger} = [d_i, d_j^{\dagger}] + d_j^{\dagger}d_i = <0|d_i d_j^{\dagger}]|0> + d_j^{\dagger}d_i $$
$$d_i d_j = <0|d_i d_j|0> + d_j d_i  $$
we move $C_1$ inside $N()$.
\begin{alignat}{2}
C_1 N(C_2 \cdots C_n) &=&& d_i (d_j^{\dagger} \cdots d_k^{\dagger} d_l \cdots d_m) \\
&=&& N(<0|C_1 C_2|0>C_3 \cdots C_n \, +\, C_2<0|C_1 C_3|0> \cdots C_n\,+ \cdots\cdots\,+\, C_2\cdots C_{n-1} <0|C_1 C_n|0> ) \\
&&&+ (d_j^{\dagger} \cdots d_k^{\dagger} d_l \cdots d_m)d_i \\
&=&& N( <0|C_1 C_2|0>C_3 \cdots C_n\, +\, C_2<0|C_1 C_3|0> \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} <0|C_1 C_n|0>\,\,\\
&&& +C_1 \cdots C_n  )
\end{alignat}
The number of $d\,\text{or}\,d^{\dagger}$ may be zero.
If $C_1=d_i^{\dagger}$, by the fact
$$ <0|d_i^{\dagger} d_j^{\dagger}|0>=0 , \quad <0|d_i^{\dagger} d_j|0>=0 , $$
no need to move $C_1$ to the right.
\begin{alignat}{2}
C_1 N(C_2 \cdots C_n) 
&=&& d_i^{\dagger} (d_j^{\dagger} \cdots d_k^{\dagger} d_l \cdots d_m) \\
&=&& N(<0|C_1 C_2|0>C_3 \cdots C_n \,+ \,C_2<0|C_1 C_3|0> \cdots C_n \,+\cdots\cdots\,+\, C_2\cdots C_{n-1} <0|C_1 C_n|0> ) \\
&&&+ d_i^{\dagger} (d_j^{\dagger} \cdots d_k^{\dagger} d_l \cdots d_m) \\
&=&& N( <0|C_1 C_2|0>C_3 \cdots C_n\, +\, C_2<0|C_1 C_3|0> \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} <0|C_1 C_n|0>\, \\
&&&+\,C_1 \cdots C_n  )
\end{alignat}
The number of $d\,\text{or}\,d^{\dagger}$ may be zero.
For the terms in which contracted $C_j$'s exist , the calculation is just in the same way since a contraction is just a multiplicative constant.
The induction step has succesfully got accomplished. 
When $C_i$'s are linear superpositions of $d_j,\,d_j^{\dagger}$, i.e.
$$ C_1=\sum_{j_1} \alpha_{j_1}C_{1j_1},\, C_2=\sum_{j_2} \alpha_{j_2}C_{2j_2}, \cdots C_n=\sum_{j_n} \alpha_{j_n}C_{nj_n} $$
Where $C_{ij_i}$ is $d_k\,\text{or}\,d_k^{\dagger}$.
Replace $ C_1\, ,\to \, C_{1j_1} \quad C_i\, \to \, \alpha_{j_i}C_{ij_i}(i\ge2)$ in the above proof.
The proof of
\begin{alignat}{2}
C_{1j_1} N(C_2 \cdots C_n) 
&=&& N( <0|C_{1 j_1} C_2|0>C_3 \cdots C_n\, +\, C_2<0|C_{1 j_1} C_3|0> \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} <0|C_{1j_1} C_n|0>\, \\
&&&+\,C_{1j_1} \cdots C_n  ).
\end{alignat}
holds as it stands for each $ C_{1j_1}=d_i,\, C_{1j_1}=d_i^{\dagger} $. 
Then multiply by $\alpha_{i_1}$, and operate $\displaystyle \sum_{i_1,i_2,\cdots,i_n}$. We'll also get
\begin{alignat}{2}
C_1 N(C_2 \cdots C_n) 
&=&& N( <0|C_1 C_2|0>C_3 \cdots C_n\, +\, C_2<0|C_1 C_3|0> \cdots C_n \,+\cdots\cdots\,+ C_2\cdots C_{n-1} <0|C_1 C_n|0>\, \\
&&&+\,C_1 \cdots C_n  ).
\end{alignat}  
And in the example, a contraction is not a commutation relation.
$$ a_2 a_3^{\dagger} = N(a_2 a_3^{\dagger} + <0|a_2 a_3^{\dagger}|0> )  $$
is obviously correct.
And since $ <0|a_1^{\dagger} a_2|0> = <0| a_1^{\dagger} a_3^{\dagger}|0>=0,$
$$ a_1^{\dagger} a_2 a_3^{\dagger} 
= N(a_1^{\dagger} a_2 a_3^{\dagger} +  <0|a_1^{\dagger} a_2|0> a_3^{\dagger} + a_2  <0| a_1^{\dagger} a_3^{\dagger}|0> + a_1^{\dagger} <0|a_2 a_3^{\dagger}|0> ) $$
holds as Wick's theorem states.
