Invertibility of a Hankel matrix Consider the following matrix of order $N+1$:
$$M(N+1) = \begin{bmatrix}
\frac{b-a}{1} & \frac{b^2-a^2}{2} & \frac{b^3-a^3}{3} & \dots & \frac{b^{N+1}-a^{N+1}}{N+1} \\
\frac{b^2-a^2}{2} & \frac{b^3-a^3}{3} & \frac{b^4-a^4}{4} & \dots & \frac{b^{N+2}-a^{N+2}}{N+2} \\
\frac{b^3-a^3}{3} & \frac{b^4-a^4}{4} & \frac{b^5-a^5}{5} & \dots & \frac{b^{N+3}-a^{N+3}}{N+3} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{b^{N+1}-a^{N+1}}{N+1} & \frac{b^{N+2}-a^{N+2}}{N+2} & \frac{b^{N+3}-a^{N+3}}{N+3} & \dots & \frac{b^{2N+1}-a^{2N+1}}{2N+1} \\
\end{bmatrix} $$
where $a \neq b$. The $(i,j)$th entry in $M(N+1)$ is $\frac{b^{i+j-1}-a^{i+j-1}}{i+j-1}$, so that it is a symmetric and Hankel matrix.
Is $M(N+1)$ invertible for any positive integer $N$? For $N = 1$ it can be easily shown that $\mathrm{det}(M(1)) = \frac{(b-a)^3}{12} \neq 0$, so $M(1)$ is invertible. In the general case I tried the obvious approach of factoring out $(b-a)$ from each entry, but I am stuck at that stage.

You may note that $$b^n-a^n=a(b^{n-1}-a^{n-1})+(b-a)b^{n-1}$$ which helps to see a relationship between rows ... then $$|M(N+1)|=(b-a)^{(N+1)^2}\times |A|,$$ where $$A_{i,j}=\frac{1}{i+j-1}.$$
