Determinant of matrix with binomial coefficient elements equal to $\binom{2n-1}{n}$ Given an $n \times n$ matrix $A_n$ with elements $a_{ij}=\binom{n+i-1}{j}$, $1 \le i \le n$, $1 \le j \le n$, I noticed that its determinant $\lvert A_n \rvert$ seem to satisfy:
$$\lvert A_n \rvert = \binom{2n-1}{n}$$
The sequence is at OEIS A001700 (shifted by $1$).
How could the identity be proved?
 A: Your determinant fits into a larger of determinants which have a very cool combinatorial interpretation, which was investigate in Binomial Determinants, Paths, and Hook Length Formulae by Ira Gessel, to which I owe all the insight in this answer.
Let
$$
A=\{a_1,a_2,\dots,a_n\},\quad B=\{b_1,b_2,\dots,b_n\}
$$
be arbitrary equal-size subsets of the nonnegative integers, sorted so that $0\le a_1<a_2<\dots<a_n$ and $0\le b_1<b_2<\dots<b_n$. Using Gessel's terminology, we define the binomial determinant for $A$ and $B$ to be the determinant of the $n\times n$ matrix whose $(i,j)$ entry is $\binom{a_i}{b_j}$. That is,
$$
\binom{a_1,\dots,a_n}{b_1,\dots,b_n}\stackrel{\text{def}}=\det\left[\binom{a_i}{b_j}\right]_{1\le i,j\le n}
$$
This reduces to your question upon setting $A=\{n,\dots,2n-1\}$ and $B=\{1,\dots,n\}$.
The following two Lemmas are useful for simplifying binomial determinants.

Lemma 1  If $b_1\neq 0$, then $$\binom{a_1,\dots,a_n}{b_1,\dots,b_n}=\frac{a_1\cdot a_2\cdots a_n}{b_1\cdot b_2\cdots b_n}\binom{a_1-1,\dots,a_n-1}{b_1-1,\dots,b_n-1}$$

Proof: Apply the absorption lemma $\binom{a_i}{b_j}=\frac{a_i}{b_j}\binom{a_i-1}{b_j-1}$ to each entry of the matrix defining $\binom{a_1,\dots,a_n}{b_1,\dots,b_n}$, then for each $i,j\in \{1,\dots,n\}$, pull out an $a_i$ from the $i^{th}$ row of the matrix and a $\frac1{b_j}$ from the $j^{th}$ column. $\;\;\;\blacksquare$

Lemma 2: $$\binom{a,a+1,\dots,a+n-1}{0,b_2,\dots,b_n}=\binom{a,a+1,\dots,a+n-2}{b_2-1,\dots,b_n-1}$$

Proof idea: This can be proven with elementary row operations. Let $M$ be the $n\times n$ matrix defining the binomial determinant on the LHS, so $M_{i,j}=\binom{a+i-1}{b_j}$, with the convention $b_1=0$. Start by subtracting the first row of $M$ from all subsequent rows. Then, subtract the second row of the result from all subsequent rows. Continue in this fashion, subtracting each row of $M$ from all subsequent rows in order from top to bottom, doing $\binom{n}2$ subtractions in all. In the final result, the first column will have a $1$ in its first entry and zeroes everywhere else. If you cofactor expand along this column, the result is exactly a smaller matrix defining the binomial determinant on the RHS. $\;\;\;\blacksquare$
Now, onto your specific case. You want to compute
$$
\binom{n,n+1,\dots,2n-1}{1,2,\dots,n}
$$
Apply the first Lemma, this equals
$$
\frac{n(n+1)\cdots(2n-1)}{1\cdot2\cdots n}\binom{n-1,n,\dots,2n-2}{0,1,\dots,n-1}
$$
Note that the fraction out front is equal to $\binom{2n-1}n$, so we just need to show the remaining binomial determinant is equal to $1$. This is accomplished by repeatedly applying the second Lemma.

There is also a combinatorial interpretation the binomial determinant in terms of collections of disjoint lattice paths. In this context, a lattice path is a path on $\mathbb Z\times \mathbb Z$ where each step is either one unit up or one unit right. Now, with sets $A$ and $B$ defined as before, define two sets of source vertices and target vertices as follows:
$$
\text{source vertices}: \quad (0,-a_1),(0,-a_2),\dots,(0,-a_n)\\
\text{target vertices}: \quad (b_1,-b_1),(b_2,-b_2),\dots,(b_n,-b_n)\\
$$
Then $\binom{a_1,\dots,a_n}{b_1,\dots,b_n}$ enumerates the number of ways to choose $n$ vertex-disjoint lattice paths, where each path connects a distinct source vertex to a distinct target vertex. The fact that this works is a direct consequence of the Lindström–Gessel–Viennot lemma.
If you apply this combinatorial interpretation to your problem, you can get a combinatorial proof that your determinant is $\binom{2n-1}n$. Specifically, when $A=\{n,\dots,2n-1\}$ and $B=\{1,\dots,n\}$, you can show the heights of the rightmost horizontal edges of the $n$ paths must form a strictly increasing sequence in order for the paths to not hit each other. Therefore, choosing the paths is equivalent to choosing a strictly increasing function $\{1,\dots,n\}\to \{1,\dots,2n-1\}$. The number of such functions is $\binom{2n-1}{n}$.
