If $B \ (\supseteq A)$ is a finitely-generated $A$-module, then $B$ is integral over $A$. I'm going through a proof of the statement:

Let $A$ and $B$ be commutative rings.
If $A \subseteq B$ and $B$ is a finitely generated $A$-module, then all $b \in B$ are integral over $A$. 

Proof:
Let $\{c_1, ... , c_n\} \subseteq B$ be a set of generators for $B$ as an $A$-module, i.e $B = \sum_{i=1}^n Ac_i$. Let $b \in B$ and write $bc_i = \sum_{j=1}^n a_{ij}c_j $ with $a_{ij} \in A$, which says that $(bI_n - (a_{ij}))c_j = 0 $ for $ 1 \leq j \leq n$. Then we must have that $\mathrm{det}(bI_n - (a_{ij})) = 0 $. This is a monic polynomial in $b$ of degree $n$. 
Why are we not done here? The proof goes on to say:
Write $1 = \alpha_1 c_1 + ... + \alpha_n c_n$, with the $\alpha_i \in A$. Then $\mathrm{det}(bI_n - (a_{ij})) = \alpha_1 (\mathrm{det}...) c_1 + \alpha_2 (\mathrm{det}...) c_2 + ... + \alpha_n (\mathrm{det}...) c_n = 0$. Hence every $b \in B$ is integral over $A$. 
I understand what is being done here on a technical level, but I don't understand why it's being done. I'd appreciate a hint/explanation. Thanks

You prematurely write "Then we must have that $\mathrm{det}(bI_n - (a_{ij})) = 0$".
At that stage you can only deduce (by multiplying by the adjoint of your matrix on the left) that all the $det\cdot c_i =0$.
However writing $1 = \alpha_1 c_1 + ... + \alpha_n c_n$ and multiplying by $det$ you do get   
$$det=det\cdot 1= \alpha_1\cdot det\cdot c_1+...+\alpha_n\cdot det\cdot c_n=\alpha_1\cdot 0+...+\alpha_n\cdot 0=0$$  
(This is a variation on  the Cayley-Hamilton theorem, according to which the characteristic polynomial of a square matrix annihilates that matrix.)
