calculus - Simplification of a nested sum

I have a nested sum like so:
$$\underbrace{\sum_{k_1=k_0}^{k^*} \ ... \sum_{k_n=k_{n-1}}^{k^*}}_{\text{n times}} 1\quad\ \text{with}\ \ n, k_0, k^* \in \mathbb{N},\ k^*\geq k_0$$

Is there a general, shorter representation that spares me calculating the actual sums?

Answer

$\displaystyle\sum_{k_0\le k_1\le\cdots\le k_n\le k^*}1=\sum_{1\le k_1-k_0+1< k_2-k_0+2<\cdots< k_n-k_0+n\le k^*-k_0+n}1={k^*-k_0+n\choose n}$