# Explicit form of a recursive relation

My question is to find an explicit formula for $$N_{k,n}$$, if the following recursive one is known:

\left\{ \begin{align} \begin{split} N_{1,n} &= 1 \text{,} \\ N_{k+1,n} &= \sum_{i = k+1}^{n-1} N_{k,i} \end{split} \end{align} \right. $$\text{for } n \in \{\,2, 3, \dots\,\}, k \in \{\,n+1, n+2, \dots\,\} \text{.}$$

I can calculate a few first terms ($$N_{1,n} = 1$$, $$N_{2,n} = n-2$$, $$N_{3,n} = \frac{(n-2)(n-3)}{2}$$), but unfortunately I cannot find the general solution.

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As mentioned in comments, one way is to notice the Hockey stick identity in the definition and then prove it matches for rest of your definition, giving the closed form $$N_{k,n}=\binom{n-2}{k-1}$$.
If you don't notice that at first, you can simplify the recurrence $$N_{k+1,n+1} - N_{k+1,n} = \sum_{i=k+1}^{n}N_{k,i}-\sum_{i=k+1}^{n-1}N_{k,i} = N_{k,n},$$ hence $$N_{k+1,n+1} = N_{k+1,n} + N_{k,n}.$$ Now add base cases such as $$N_{1,n}=1$$ and $$N_{k,2}=0$$ for $$k\geq 2$$, which follow trivially from definitions. Again you can notice that the recurrence definition resembles Pascal's formula and you can go from there.
Eventually if all that fails, you can use a generic method for solving the last recurrence. For example, using generating functions as in the answer Solving two-dimensional recurrence relation $a_{i,\ j}\ =\ a_{i,\ j-1}\ +\ a_{i-1,\ j-1}$, you should get $$f(x,y)=\sum_{k\geq 1, n \geq 2}N_{k,n}x^k y^n=\frac{xy^2}{1-y-xy}$$. This in turn simplifies using geometric series into $$f(x,y)=xy^2\sum_{n=0}^{\infty}y^n(1+x)^n=xy^2\sum_{n=0}^{\infty}y^n\sum_{k=0}^{n}\binom{n}{k}x^k = \sum_{n\geq 0, k \geq 0}^{\infty}\binom{n}{k}x^{k+1}y^{n+2},$$ and so finally $$f(x,y)=\sum_{n\geq 2, k \geq 1}^{\infty}\binom{n-2}{k-1}x^{k}y^{n}.$$ Now just read off the coefficient of $$x^k y^n$$.