<s>[INST] Can we determine the determinant? Could someone prove that this determinant is not zero?
$$\left|
      \begin{array}{cccc}
        1^n & 2^n & \cdots & (n+1)^n \\
        2^n & 3^n & \cdots & (n+2)^n \\
        \vdots & \vdots &  & \vdots \\
        (n+1)^n & (n+2)^n & \cdots & (2n+1)^n \\
      \end{array}
    \right|\neq0$$
How can we compute it?
 [/INST] First we apply binomial theorem, e.g. write $2^n$ as $(1+1)^n$, then $2^n=1^n+C_n ^1 1^{n-1}1^1+ \cdots+C_n ^n 1^0 1^n$. And note that $\det(AB)=\det(A)\det(B)$ for square matrix $A,B$ with same order.
$$\left|
      \begin{array}{cccc}
        1^n & 2^n & \cdots & (n+1)^n \\
        2^n & 3^n & \cdots & (n+2)^n \\
        \vdots & \vdots &  & \vdots \\
        (n+1)^n & (n+2)^n & \cdots & (2n+1)^n \\
      \end{array}
    \right|
=
\left|
      \begin{array}{cccc}
        (0+1)^n & (0+2)^n & \cdots & (0+(n+1))^n \\
        (1+1)^n & (1+2)^n & \cdots & (1+(n+1))^n \\
        \vdots & \vdots &  & \vdots \\
        (n+1)^n & (n+2)^n & \cdots & (n+(n+1))^n \\
      \end{array}
    \right|
=
\left|
      \begin{array}{cccc}
        0 & 0 & \cdots & 1 \\
        1^n & 1^{n-1} & \cdots & 1^0 \\
        \vdots & \vdots &  & \vdots \\
        n^n & n^{n-1} & \cdots & n^0 \\
      \end{array}
    \right| \cdot 
\left|
      \begin{array}{cccc}
        1 & 1 & \cdots & 1 \\
        C_n^1 1^1 & C_n^1 2^{1} & \cdots & C_n ^1 (n+1)^1 \\
        \vdots & \vdots &  & \vdots \\
        C_n^n 1^n & C_n^n 2^{n} & \cdots & C_n^n (n+1)^n \\
      \end{array}
    \right|
$$
Notice that this 2 matrix are nearly of the Vandermonde form. See here. Except that the first one should have all its columns reversed; the second one should extract the factors $C_n ^1, \dots,C_n ^n$. Hence,
$$
\left|
      \begin{array}{cccc}
        0 & 0 & \cdots & 1 \\
        1^n & 1^{n-1} & \cdots & 1^0 \\
        \vdots & \vdots &  & \vdots \\
        n^n & n^{n-1} & \cdots & n^0 \\
      \end{array}
    \right| \cdot 
\left|
      \begin{array}{cccc}
        1 & 1 & \cdots & 1 \\
        C_n^1 1^1 & C_n^1 2^{1} & \cdots & C_n ^1 (n+1)^1 \\
        \vdots & \vdots &  & \vdots \\
        C_n^n 1^n & C_n^n 2^{n} & \cdots & C_n^n (n+1)^n \\
      \end{array}
    \right|= \\
(-1)^{n(n+1)/2}C_n ^1 C_n ^2 \cdots C_n ^n 
\left|
      \begin{array}{cccc}
        1 & 0 & \cdots & 0 \\
        1 & 1^{1} & \cdots & 1^n \\
        \vdots & \vdots &  & \vdots \\
        1 & n^{1} & \cdots & n^n \\
      \end{array}
    \right| \cdot 
\left|
      \begin{array}{cccc}
        1 & 1 & \cdots & 1 \\
         1^1 &  2^{1} & \cdots &   (n+1)^1 \\
        \vdots & \vdots &  & \vdots \\
         1^n &  2^{n} & \cdots &  (n+1)^n \\
      \end{array}
    \right|
\\ =(-1)^{n(n+1)/2}C_n ^1 C_n ^2 \cdots C_n ^n \prod_{0 \leq i <j \leq n}(j-i) \prod_{1 \leq i <j \leq n+1}(j-i)
\neq 0.
$$
</s>