proof of Nakayama's lemma Let $R$ a local ring with maximal ideal $\mathfrak m$. Suppose $M$ is finitely generated $R-$module s.t. $\mathfrak m M=M$. Prove that $M=0$.
Proof : Let $m_1,...,m_n$ be generator of $M$. As $$\mathfrak m M=M,$$
there is a matrix $A$ with entries in $\mathfrak m$ s.t. $$A\bar m=\bar m$$
where $\bar m=(m_1,...,m_n)^T$. 
Q1) Why is there such a matrix ? Where does it come from ?
Multiplying by the adjugate of the matrix $A-I$ implies that if $D=\det(A-I)$, then $Dm_i=0$ for all $i$.
Q2) I don't understand what do they mean by "multipliying by adjugate of the matrix $A-I$". I saw on wikipedia what is the adjugate, but it doesn't help.
By expanding the determinant, we can see that $D=\det (A-I)$ cannot be in $\mathfrak m$, for all the component of $A$ are in $\mathfrak m$, and so, there is only one term of the expansion of $\det(A-I)$ which is not, and this is the 1 which come from the $\prod(a_{ii}-1)$ terms. 
Q3) I still don't understand what they mean, it looks really confuse...
It's not necessary to put the rest, since here I understand nothing of what they do. If it's clear for someone, I would appriciate some explanations...
 A: The matrix exists because of the condition $\mathfrak{m}M=M$. That means that there are $a_{i1},a_{i2},\ldots,a_{in}\in\mathfrak{m}$ such that 
$$\sum_{j=1}^n{a_{ij}m_j}=m_i$$
for all $i$. This is the same thing as the existence of such a matrix.
The adjugate $B'$ matrix of a matrix $B$ has the property that 
$$BB'=\det(B)I$$
where $I$ is the identity matrix. This is what is important here. Since 
$$(A-I)\overline{m}=0$$
we must have that $Dm_i=0$ for all $i$.
What they mean for Q3 is that if you take the formula for the determinant as a sum 
$$\sum_{\sigma\in S_n}{(-1)^{\mathrm{sgn}(\sigma)}b_{1\sigma(1)}\cdots b_{n\sigma(n)}}$$
then since the diagonal entries are not in $\mathfrak{m}$ but every other entry is we must have that every term in is $\mathfrak{m}$ (because $\mathfrak{m}$ is an ideal) except for the term where $\sigma$ is the identity, when none of the elements of the product are in $\mathfrak{m}$; this means that $D+\mathfrak{m}\neq\mathfrak{m}$, so $D\notin\mathfrak{m}$.
A: Matt Samuel's answers explains pretty well the situation.
However, since Nakayama's lemma holds in a much more general setting, it could help your intuition if you see a different proof.

Let $R$ be a ring (not necessarily commutative, but with identity). Let $J(R)$ be its Jacobson radical and suppose $M$ is a finitely generated left $R$-module with $M=J(R)M$. Then $M=0$.

The Jacobson radical is the intersection of all maximal left ideals (it is the same as the intersection of all maximal right ideals). The property we need is that

if $r\in J(R)$, then $1-r$ is left invertible

which is rather easy to show: if $R(1-r)\ne R$, then $R(1-r)$ is contained in a maximal left ideal $I$; on the other hand $r\in J(R)$ implies $r\in I$, so $1=(1-r)+r\in I$: contradiction.
Let $\{x_1,\dots,x_k\}$ be a minimal generating set for $M$, in the sense that no proper subset of it generates $M$; assume $k\ge1$. Since $M=J(R)M$, we can write
$$
x_k=\sum_{i=1}^k r_ix_i
$$
with $r_i\in J(R)$ and, in particular
$$
(1-r_k)x_k=\sum_{i=1}^{k-1}r_ix_i
$$
Take a left inverse $s$ of $1-r_k$; then
$$
x_k=\sum_{i=1}^{k-1}sr_ix_i
$$
contradicting the mimimality of $\{x_1,\dots,x_k\}$. QED
Such a proof is not “constructive”, as it relies on existence of maximal ideals. However, it can be very easily adapted to your situation and it doesn't require any determinant based argument.

The last part can be explained more easily, too. If you look at $A-I$ modulo $\mathfrak{m}$, then it is $-I$, so its determinant modulo $\mathfrak{m}$ is $(-1)^n$. Therefore the determinant of $A-I$ does not belong to $\mathfrak{m}$, so it is invertible in $R$
