How can I calculate this determinant? Please can you give me some hints to deal with this :
$\displaystyle \text{Let } a_1, a_2, ..., a_n \in \mathbb{R}$
$\displaystyle \text{ Calculate } \det A \text{ where }$ $\displaystyle A=(a_{ij})_{1\leqslant i,j\leqslant n} \text{ and }$ $\displaystyle \lbrace_{\alpha_{ij}=0,\text{ otherwise}}^{\alpha_{ij}=a_i,\text{ for } i+j=n+1}$

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 
$$\det A= (-1)^{\frac{n(n-1)}{2}}a_1\times \cdots\times a_n$$
