Closed form of recurrent arithmetic series summation

We can write the last multiple sum as \begin{align*} \color{blue}{\sum_{i_1=1}^n\sum_{i_2=1}^{i_1}\sum_{i_3=1}^{i_2}i_3} &=\sum_{i_1=1}^n\sum_{i_2=1}^{i_1}\sum_{i_3=1}^{i_2}\sum_{i_4=1}^{i_3} 1\\ &=\sum_{1\leq i_4\leq i_3\leq i_2\leq i_1\leq n}1\tag{1}\\ &\,\,\color{blue}{=\binom{n+3}{4}}\tag{2} \end{align*} In (1) we observe the index range is the number of ordered $4$-tuples with repetition from a set with $n$ elements resulting in (2).

Let $f_k(n)$ be the closed form of the summation nested $k$ times. We know that $$f_1(n)=\frac12n(n+1)=\binom{n+1}{2}$$ $$f_k(n)=\sum_{j=1}^n f_{k-1}(j)$$ So for the next function $f_2(n)$ we have $$f_2(n)=\sum_{j=1}^n\binom{j+1}{2}=\sum_{j=2}^{n+1}\binom{j}{2}=\binom{n+2}{3}$$ By using the Hockey-stick identity (credits to Jean-Claude Arbaut). Similarly for the next function $f_3(n)$ we have $$f_3(n)=\sum_{j=1}^n\binom{j+2}{3}=\sum_{j=3}^{n+2}\binom{j}{3}=\binom{n+3}{4}$$ So one could conjecture that $$f_k(n)=\binom{n+k}{k+1}$$ which can be easily proven by induction as follows $$f_k(n)=\sum_{j=1}^n\binom{j+k-1}{k}=\sum_{j=k}^{n+k-1}\binom{j}{k}=\binom{n+k}{k+1}$$ Hence we have that $$\boxed{f_k(n)=\binom{n+k}{k+1}=\frac1{(k+1)!}n(n+1)(n+2)\dots(n+k-1)(n+k)}$$