How can I calculate this determinant? Please can you give me some hints to deal with this :
$\displaystyle \text{Let } a_1, a_2, ..., a_n \in \mathbb{R}$
$\displaystyle \text{ Calculate } \det A \text{ where }$ $\displaystyle A=(a_{ij})_{1\leqslant i,j\leqslant n} \text{ and }$ $\displaystyle \lbrace_{\alpha_{ij}=0,\text{ otherwise}}^{\alpha_{ij}=a_i,\text{ for } i+j=n+1}$
 A: Hint: The matrix looks like the following (for $n=4$; it gives the idea though):
$$
\begin{bmatrix}
0 & 0 & 0 & a_1\\
0 & 0 & a_2 & 0\\
0 & a_3 & 0 & 0\\
a_4 & 0 & 0 & 0
\end{bmatrix}
$$
What happens if you do a cofactor expansion in the first column? Try using induction.
A: Recall that the determinant of the matrix  $\displaystyle A=(a_{ij})_{1\leqslant i,j\leqslant n}$ is given by
$$\det A=\sum_{\sigma\in S_n}\epsilon(\sigma)\prod_{i=1}^n a_{i\sigma(i)}$$
and by the form of the matrix $A$ the only permutation that gives a non zero term is
$$\sigma=\left(\begin{matrix}1&2&\cdots&n\\
n&n-1&\cdots&1\end{matrix}\right)$$
and its signature is the number of inversion so 
$$\epsilon(\sigma)=(-1)^{(n-1)+(n-2)+\cdots+1}=(-1)^{\frac{n(n-1)}{2}}$$
hence we conclude that 
$$\det A= (-1)^{\frac{n(n-1)}{2}}a_1\times \cdots\times a_n$$
A: As @MarianoSuárez-Alvarez said, write down the matrix for small $n$ and see how it looks like.
Your matrix is just an anti-diagonal matrix, whose determinant is given by $(-1)^{\frac{n(n-1)}{2}}\prod_{i=1}^n a_i$. 
The sign part $(-1)^{\frac{n(n-1)}{2}}$ comes from arranging the rows to achieve a diagonal matrix whose determinant is just the product of all elements on the diagonal. Swapping two adjacent lines will change the sign of the determinant.
