Define the companion family of finite sequences $\{Q_{ni}\}$ such that $Q_{n1}=(1)$, $Q_{nn}=(n)$ and $Q_{ni}=Q_{(n-1)(i-1)}Q_{(n-1)i}$ whenever $1<i<n$. Claim 1. $Q_{ni}$ has length $\binom {n-1}{i-1}$. Proof. This is clear for $i=1$ and $i=n$. Using induction, we find that $Q_{(n+1)(i+1)}=Q_{ni}Q_{n(i+1)}$ has length \[\binom{n-1}{i-1}+\binom{n-1}i=\binom ni,\] as needed. $\blacksquare$ Now define for any finite sequence $A=(x_1,x_2,\dots, x_n)$ (where all the $x_i$ are positive integers) the following operation: \[\hat{A}=(1,2,\dots,x_1,1,2,\dots,x_2,\dots,1,2,\dots,x_n).\] Then of course we have $\hat{A}\hat{B}=\hat{AB}$. Claim 2. $Q_{ni}=\hat{Q_{(n-1)i}}$ for $n\ge 2$ and $1\le i\le n-1$. Proof. We use induction on $n$. This is trivial for $i=1$ because $Q_{n1}=Q_{(n-1)i}=(1)$. This also implies the truth of the claim for $n=2$. For $n>2$, $1<i<n$, we find by the induction hypothesis \[Q_{ni}=Q_{(n-1)(i-1)}Q_{(n-1)i}=\hat{Q_{(n-2)(i-1)}}\hat{Q_{(n-2)i}}=\hat{Q_{(n-1)i}}.\blacksquare\] Claim 3. The $j$-th term from the left in $Q_{ni}$ is a $1$ if and only if the $j$-th term from the right in $Q_{n(n+1-i)}$ is not a $1$ ($n\ge 2$). Proof. Inducting on $n$, starting from the trivial $n=2$, we note that either $i=1$ or $i=n$, in which case we are done based on the definition, or else we have $Q_{ni}=Q_{(n-1)(i-1)}Q_{(n-1)i}$ and $Q_{n(n+1-i)}=Q_{(n-1)(n-i)}Q_{(n-1)(n+1-i)}$. In the latter case, the $j$-th term from the left in $Q_{ni}$ is either in the part $Q_{(n-1)(i-1)}$, or it is in $Q_{(n-1)i}$. Because of Claim 1, we notice that $Q_{(n-1)(n+1-i)}$ is just as long as $Q_{(n-1)(i-1)}$, so the $j$-th term from the left in $Q_{ni}$ and the $j$-th term from the right in $Q_{n(n+1-i)}$ coincide with either the $j$-th term from the left in $Q_{(n-1)(i-1)}$ and from the right in $Q_{(n-1)(n+i-1)}$, or the $j^*$-th term (for some $j^*$) from the left in $Q_{(n-1)i}$ and from the right in $Q_{(n-1)(n-i)}$. In both cases, we are done because of the induction hypothesis. $\blacksquare$ Claim 4. $R_n=Q_{n1}Q_{n2}\dots Q_{nn}$. Proof. This is trivial for $n=1$. Proceeding by induction and using the definition of $R_n$ and also Claim 2, we see that \[R_n=\hat{R_{n-1}}Q_{nn}=\hat{Q_{(n-1)1}}\hat{Q_{(n-1)2}}\dots\hat{Q_{(n-1)(n-1)}}Q_{nn}=\] \[=Q_{n1}Q_{n2}\dots Q_{n(n-1)}Q_{nn}.\blacksquare\] Now take the $k$-th term from the right in $R_n$. Using the identity of Claim 4, we see that this is inside some sequence $Q_{ni}$, at place number $j$ (from the left). Because of Claim 1, this implies \[k=\binom {n-1}0+\binom{n-1}1+\dots+\binom{n-1}{i-2}+j=\] \[=\binom{n-1}{n-1}+\binom{n-1}{n-2}+\dots+\binom{n-1}{n+1-i}+j.\] Hence, using Claim 1 again, we find that the $k$-th term from the right in $R_n$ is none other than the $j$-th term from the right in $Q_{n(n+1-i)}$. By Claim 3, this implies the statement of the problem.