Is there any approximation for
$$\sum\limits_{i_3=3}^{n-k+3}\sum\limits_{i_4=i_3+1}^{n-k+4}\sum\limits_{i_5=i_4+1}^{n-k+5}\cdots\sum\limits_{i_k=i_{k-1}+1}^{n} 1, \quad (n \gg k)$$ ?
We know that $\sum\limits_{i_3=3}^{n-k+3}\sum\limits_{i_4=i_3+1}^{n-k+4}\sum\limits_{i_5=i_4+1}^{n-k+5}\cdots\sum\limits_{i_k=i_{k-1}+1}^{n} 1 < \sum\limits_{i_3=3}^{n-k+3}\sum\limits_{i_4=4}^{n-k+4}\sum\limits_{i_5=5}^{n-k+5}\cdots\sum\limits_{i_k=k}^{n} 1 = (n-k+1)^{k-2}$
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