Vtyjkyuk

Let $ninmathbbN$. I observed today that
$$1+n+sum_i_1=1^ni_1+sum_i_2=1^nsum_i_1=1^i_2i_1+sum_i_3=1^nsum_i_2=1^i_3sum_i_1=1^i_2i_1+dots + sum_i_n-1^nsum_i_n-2=1^i_n-1dotssum_i_1=1^i_2i_1=frac(2n)!(n!)^2$$
I'm wondering how this identity can be proven. Note that there are $n+1$ terms on the LHS.

Ignoring the initial $1$ on the LHS, you can write the $k^th$ term on the LHS as
$$
sum_i_1=1^nsum_i_2=1^i_1cdots sum_i_k=1^i_k-1i_k=sum_i_1=1^nsum_i_2=1^i_1cdots sum_i_k=1^i_k-1sum_i_k+1=1^i_k1
$$
Since we are summing the number $1$, the value of the summation is just equal to the number of ways to choose the indices.

A choice of indices is a weakly decreasing list $nge i_1ge i_2ge dots ge i_kge i_k+1ge 1$. Choosing these is equivalent to choosing a multi-set of $[n]$ of size $k+1$, the number of which is $binomn+kn-1$. Therefore, the LHS is
$$
1+binomnn-1+binomn+1n-1+dots+binom2n-1n-1,
$$
and the result follows by the hockey stick identity.

Fix $1leqslant pleqslant 2n$ and let $A_p$ denote the number of ways to choose an $n$ element subset $T$ of $1, ldots, 2n$ with $max T=p$. Note that $A_p=0$ for $p<n$, $A_n=1$, and for $1leqslant jleqslant n$, $$A_n+j=sum_i_j=1^nldots sum_i_2=1^i_3 sum_i_1=1^i_21 = sum_i_j=1^nldots sum_i_2=1^i_3i_2.$$ Summing over $p$ gives the number of $n$ element subsets of $2n$, which is $binom2nn$, your right side, and it is also equal to $sum_p=n^2nA_p$, which is your left side.