Determinant of a symmetric matrix of order $2018$ 
Find out the determinant of the following matrix :
$$\begin{bmatrix}1^{2016} & 2^{2016} & ... & 2018^{2016}\\2^{2016} & 3^{2016} & ... & 2019^{2016}\\.. & .. & ... & ..\\2018^{2016} & 2019^{2016} & ... & 4035^{2016}\end{bmatrix}$$

Through examples of order $2\times 2$ as $\displaystyle \begin{bmatrix}1^0 & 2^0\\2^0 & 3^0\end{bmatrix}$ and $4\times 4$ as $\displaystyle \begin{bmatrix}1^2 & 2^2 & 3^2 & 4^2\\2^2 & 3^2 & 4^2 & 5^2\\3^2 & 4^2 & 5^2 & 6^2\\4^2 & 5^2 & 6^2 & 7^2\end{bmatrix}$, I found that in both case the answer is  $0$. But I want to know the procedure to find such determinant.
 A: So, what we do is the following : The above matrix $M$, seems to be given by $M_{ij} = (i+j-1)^{2016}$.
Expanding this binomially, we get $$M_{ij} = \sum_{x =  0}^{2016} \binom {2016}x
(i-1)^{2016-x} j^{x}$$
Now, define the $2017$ column vectors $C_a$, $0 \leq a \leq 2016$, by $C_a = [1^a,2^a,...,2018^a]^T$. I claim that every column of $M_{ij}$ is spanned by these column vectors.
This happens, because given a column $M_{i}$, we can see from the above formula that:
$$
M_i = \sum_{x=0}^{2016}\left(\binom{2016}{x}(i-1)^{2016-x}\right)C_x
$$
For every $i$. I urge you to check this yourself.
This shows that $M$ has rank $2017$, since $0 \leq a \leq 2016$. However, the matrix you have given is square of dimension $2018$. It follows that it has non-trivial kernel, hence is not injective, hence has determinant zero.
A: Suppose  $p_1, p_2...,p_n $ are polynomials  of degree $m-1 $  with atleast one has degree exactly $m-1$ . Let  us consider  a matrix $A$  whose  $(i,j)-$th matrix has  entry  $p_i(u_j)$ , where $u_1,u_2,.......,u_n$ are $ n$ distinct  number 
let $p_i(x) = a_{i,1} + a_{i,2} x +.....+a_{i,m}x^{m-1} $
Now  we can easily show  that $A=\displaystyle \begin{bmatrix}a_{1,1} & a_{1,2} .......&a_{1,m}\\ . & ....... & .\\. & ....... & .\\a_{n,1} & a_{1,2} .......&a_{n,m}\end{bmatrix}\displaystyle \begin{bmatrix}1 & 1 & ..... & 1\\u_1 & u_2 & ..... & u_n\\. & .& ..... & .\\. & .& ..... & .\\u_1^{m-1} & u_2^{m-1} & .... & u_n^{m-1}\end{bmatrix}$,
This  show that in particular  if  max $deg(p_i) = m_i < n$ , (this mean , the  first matrix  is of size  $n \times m$ and$ m < n$ ) , then $rank (A) < n$  and so determinant is $0$
For the given matrix  $p_i(x) = (i-1 +x)^{2016}$ and points  are $ 1,....., 2018$. Since the matrix is of size   $2018 \times 2018$, the determinant is $0$
