Proving that $\det(A) \ne 0$ if $a_{i,i} = 0$ and $a_{i,j} = \pm 1$ for $i \neq j$ 
Let $A$ be an $n\times n$ matrix ($n=2k$, $k \in \Bbb N^*$) such that.
$$a_{ij} =
\begin{cases}
\pm 1,  & \text{if $i \ne j$} \\
0, & \text{if $i=j$}
\end{cases}$$ 
Show that $\det (A) \ne 0$.

P.S. $a_{ij}=\pm 1$ means that it can be $+1$ or $-1$ not necessarily the same for all $a_{ij}$. 
My approach:
I've started with the definition of $\det A$ writing like a permutation sum but it became messy. I also tried Laplace's method but also didn't work. I also tried induction but once $\pm 1$ is aleatory it became tough to deal with.
 A: Without finite fields: We wish to show that $\det(A)$ is necessarily odd (which we concluded from the $\Bbb F_2$ approach), which would imply $\det(A) \neq 0$.  Fix a matrix $A$ of the described pattern.
Let $A^{(i,j)}$ denote the matrix in which we "flip" the sign of the $i,j$ entry.  That is,
$$
A^{(i,j)}_{pq} = \begin{cases}
-A_{ij} & p=i,q=j\\
A_{ij} & \text{otherwise}
\end{cases}
$$
By applying the Laplace expansion of the determinant along the $i$th row, we see
$$
\det(A) - \det(A^{(i,j)}) = 2C_{ij}
$$
Where $C_{ij}$ is the cofactor associated with $i,j$.  Notably, $C_{ij}$ is an integer.
So, flipping one sign won't change the parity of the derivative.  Thus, is sufficient to show that the matrix $A$ whose off-diagonal entries are all $1$ has an odd determinant.  But this is easy.  
In particular, if $x$ is the column vector whose entries are all $1$, then we want to show that $\det(xx^T - I)$ is odd.  We can do so either by considering the eigenvalues of the rank-$1$ matrix $xx^T$ or by Sylvester's determinant identity (or by row-reduction, or by observing that the matrix is circulant).  Whichever way you choose, we find $\det(xx^T - I) = 1-n$, which is necessarily odd.

Another slick proof using finite fields (which avoids appealing to circulant matrices): 
Consider the $n \times n$ matrix $B$ whose entries (from $\Bbb F_2$) are all $1$.  Since $n$ is even, we find that $B^2 = nB = 0$. Since $B$ is nilpotent, its only eigenvalue is $0$ (with multiplicity $n$).
The matrix $A$ (taken modulo $2$) is given by $A = B + I$.  Since $B$ has eigenvalue $0$ with multiplicity $n$, we may conclude that $A$ has eigenvalue $1$ with multiplicity $n$. Thus, $\det(A) = (1)^n = 1 \neq 0$.
Thus, we conclude that the determinant of an $A$ of the presented form is odd, hence non-zero.

Or, more simply: $B^2 = 0$.  So, $A = B+I$ is invertible since 
$$
A^2 = B^2 + 2B + I = 0 + 0 +I = I
$$
A: Thank you @Pierre-Guy, your solution and particularly the matrix $B$ gave me an idea.
If we write $\det A$ using congruence modulo $2$, we get that $\pm 1$ will became $1$ and then we have:
$$\det A \equiv\begin{vmatrix}
        0 & 1 & 1 & ...&1 \\
        1 & 0 & 1 & ...&1 \\
        \vdots & \vdots & \vdots & ...&1 \\
        1 & 1 & 1 & ...&0
\end{vmatrix}$$
and sum every horizontal line at the first we have:
$$\begin{vmatrix}
        2k-1 & 2k-1 & 2k-1 & ...&2k-1 \\
        1 & 0 & 1 & ...&1 \\
        \vdots & \vdots & \vdots & ...&1 \\
        1 & 1 & 1 & ...&0
\end{vmatrix}=(2k-1)\begin{vmatrix}
        1 & 1 & 1 & ...&1 \\
        1 & 0 & 1 & ...&1 \\
        \vdots & \vdots & \vdots & ...&1 \\
        1 & 1 & 1 & ...&0
\end{vmatrix}$$
Now if we subtract every horizontal line from the first line we get:
$$(2k-1)\begin{vmatrix}
        1 & 1 & 1 & ...&1 \\
        0 & -1 & 1 & ...&1 \\
        \vdots & \vdots & \vdots & ...&1 \\
        0 & 0 & 0 & ...&-1
\end{vmatrix}=(2k-1)(-1)^{(2k-1)}\equiv 1 \mod 2$$
and so the $\det A$ is a odd number and then it can't be $0$.
A: Here's an approach which connects to your original idea of writing out the definition of $\det A$ as a sum over permutations:
If you to this, you get $n!$ terms, each of which is either $+1$, $-1$ or $0$. The terms which are zero are the ones where you include at least one matrix entry from the diagonal, so they correspond exactly to the permutations $\pi$ which have at least one fixpoint (i.e., $\pi(k)=k$ for some $k$).
So the nonzero terms in the sum correspond to the fixpoint-free permutations, also known as derangements. And the number of derangements of $n$ elements, call it $a_n$, is odd if $n$ is even (and even if $n$ is odd), which is easy to prove from the recursion
$$
a_1=0
,\qquad
a_n = n a_{n-1} + (-1)^n
.
$$
So your determinant is the sum of an odd number of $+1$ and $-1$, hence it is itself an odd number, and can't be zero.
A: Assume that $n$ is even. Consider your matrix $A$ as being a matrix with integer coefficient, and let $\mathbb{F}_2$ be the field with two elements.  Then the projection $\pi:\mathbb{Z}\to \mathbb{F}_2$ induces a projection $\pi$ on the rings of matrices; let $A_2$ be the image of $A$ by this projection.  Then $\det(A_2) = \pi(\det(A))$.
Now, let $B$ be the matrix defined in the same way as $A$, except that all non-zero entries are $1$ (instead of $\pm 1$).  Then $\pi(B)= A_2$.  Now $B$ is a circulant matrix, and we know that its determinant is $(n-1)\cdot(-1)^n$.  Hence $\det(A_2)=1$, so $\det(A)$ cannot be zero (in fact, it cannot be an even number).
Note that this also shows that if $n$ is odd, then $\det(A)$ will always be an even number.

Added:
After seeing Omnomnomnom's answer, I realized that the matrix $A_2$ above is invertible, and is its own inverse, if $n$ is even.  So there's no need to use the circulant matrix $B$ in the proof.

And if $n$ is odd:
The result is false if $n$ is odd.  A counter-example for $n=3$ is given by the matrix
$$
\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & -1 & 0
\end{pmatrix}.
$$
If $n$ is any odd integer, then this counter-example generalizes: let $A$ have $1$'s everywhere, except on the diagonal, and also on the last row, where $1$ and $-1$ alternate.  Then $A$ has zero determinant, since the vector $(1,1,\ldots,1,n-2)^t$ is in its kernel.
