= Solution
The displayed identity is valid for a <highest-weight vector> $v$, satisfying $Xv=0$ and $Hv=kv$, not for arbitrary $v\in V$. For example, in the defining module with $k=1$, take $v=y$: then $XYv=0$ but the proposed right side for $n=1$ is $v\ne0$.
For a <highest-weight vector>, the <sl2 triple> relations give $HY^jv=(k-2j)Y^jv$. The operator identity
$$
[X,Y^n]=\sum_{j=0}^{n-1}Y^jHY^{n-1-j}
$$
comes from expanding a <commutator> with a product. Apply it to $v$, using $Xv=0$. Each term becomes $(k-2(n-1-j))Y^{n-1}v$, so the sum is
$$
\boxed{XY^nv=n(k-n+1)Y^{n-1}v.}
$$
This proves the intended <sl2 highest-weight lowering formula>. The identity valid for an arbitrary vector is instead
$$
XY^nv=Y^nXv+nY^{n-1}(H-n+1)v.
$$
On the <basis> $v,Yv,\ldots,Y^kv$ of the irreducible module, $XY$ has <eigenvalue> $(j+1)(k-j)$ on $Y^jv$, for $0\leq j\leq k$. Summing gives the <trace of a raising-lowering product in an irreducible sl2 module>:
$$
\boxed{\operatorname{tr}_V(XY)=\sum_{j=0}^k(j+1)(k-j)=\frac{k(k+1)(k+2)}6.}
$$
The last equality follows by the formulas for the sums of the first $k$ integers and their squares.
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