@Tsemo Aristide is absolutely correct, you can follow that link and also The Proof for your specific case here. However, this is a different kind of explanation for what you have, which is not a proof but I think it might help you better grasp the concept.
At the first glance, this looks very similar to the Vandermonde matrix; except we have
$$ M = \left[ {\begin{array}{ccccc}
x_1 & x_2 & x_3 & \cdots & x_n
\\x_1^2 & x_2^2 & x_3^2 &\cdots & x_n^2\\
\vdots & & \ddots & & \vdots \\
x_1 ^n & x_2^n & x_3^n & \cdots &x_n^n \\
\end{array} } \right]$$
Now let us consider the Vandermonde's matrix which is
$$V = \left[ {\begin{array}{ccccc}
1 & x_1 & x_1^2& \cdots & x_1^{n-1}\\
1 & x_2 & x_2^2 &\cdots & x_2^{n-1}\\
\vdots & & \ddots & & \vdots \\
1 & x_n& x_n & \cdots &x_n^{n-1}\\
\end{array} } \right]$$
As we can see, there are very closely related; so we want to find the matrix $X$ so that
$$VX=M$$ and in that case the determinant of $M$ will be
$$\det(M) = \det(V) \cdot \det(X)$$
Now let us look at an example of a $2 \times 2$ by $2 \times 2$ matrices, so assume that we have :
$$\left[ {\begin{array}{cc}
a & 0\\
0 & b \\
\end{array} } \right]
\times
\left[ {\begin{array}{cc}
1 & 1\\
2& 2 \\
\end{array} } \right]
=
\left[ {\begin{array}{cc}
a & a\\
2b& 2b \\
\end{array} } \right]
$$
So we can see that by multiplying this diagonal matrix by a the square matrix, we can multiply each component of the first row by $a$ and second row by $b$. Now this is exactly what we want in order to go from our the Vandermonde matrix to the given matrix; furthermore this is very beneficial since we know the determinant of the diagonal matrix is the multiplication of the diagonals and we know the Vandermonde's determinant. So to get $M$ we will have
$VX = N $ where $$X =
\left[ {\begin{array}{ccccc}
x_1 & 0 & 0& \cdots & 0\\
0 & x_2 & 0 &\cdots & 0\\
\vdots & & \ddots & & \vdots \\
0&0 &0 & \cdots &x_n\\
\end{array} } \right]
$$
Now
$$VX = \left[ {\begin{array}{ccccc}
x_1 & x_1^2 & x_1^3 & \cdots & x_1^n\\
x_2 & x_2^2 & x_2^2 &\cdots & x_2^n\\
\vdots & & \ddots & & \vdots \\
x_n & x_n^2 & x_n^3 & \cdots &x_n^n\\
\end{array} } \right] = N
$$
This is almost what we want except we have the transpose of this matrix, so $(VX)^T = M$. Also, for a square matrix, the determinant of the transpose is the same as the determinant of the matrix, we will have :
$$\det(X)\cdot\det(V) = \prod_{1\leq i \leq n}x_i \cdot \prod_{1 \leq i \leq j \leq n} (x_j - x_i) = \det(M^T) = \det(M)$$
Again this is not a proof, but I hope this clarified some stuff from both of the links.