Let $L_i$ and $R_i$ be the $i$-th rows of the left and right matrices. Then $$\begin{align} R_n &= L_n \\ R_{n-1} &= L_n + L_{n-1} \\ R_{n-2} &= L_n + 2L_{n-1} + L_{n-2} \\ R_{n-3} &= L_n + 3L_{n-1} + 3L_{n-2} + L_{n-3} \\ R_{n-4} &= L_n + 4L_{n-1} + 6L_{n-2} + 4L_{n-3} + L_{n-4}, \end{align}$$ and so forth, with the coefficients being binomial coefficients. Each sum is just a Vandermonde convolution.

So the second matrix can be derived from the first by elementary row operations, which we know preserve the determinant.

Let $L_i$ and $R_i$ be the $i$-th rows of the left and right matrices. Then $$\begin{align} R_n &= L_n \\ R_{n-1} &= L_n + L_{n-1} \\ R_{n-2} &= L_n + 2L_{n-1} + L_{n-2} \\ R_{n-3} &= L_n + 3L_{n-1} + 3L_{n-2} + L_{n-3} \\ R_{n-4} &= L_n + 4L_{n-1} + 6L_{n-2} + 4L_{n-3} + L_{n-4}, \end{align}$$ and so forth, with the coefficients being binomial coefficients. Each sum is just a Vandermonde convolution.

So the second matrix can be derived from the first by elementary row operations, which we know preserve the determinant. (The simplest way is to work from the first row downwards, adding the required multiples of the rows below.)