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.

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.
