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{\bf Solution of 12022. (Proposed by Mircea Merca), by Shalosh B. Ekhad, \hfill\break ShaloshBEkhad@gmail.com}

Let $F(n,k)$ be the summand. We have to prove that $a(n):=\sum_{k=0}^{n-1} F(n,k)$ equals $n$.
We cleverly come up with $G(n,k) \,= \, -\,(1-x^n) \cdots (1-x^{n-k+1})$, that satisfies \hfill\break
$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$ (check!) . Hence 
$$
a(n+1)-a(n)=F(n+1,n)+ \sum_{k=0}^{n-1} ( \, F(n+1,k) -  F(n,k)\, )=
F(n+1,n)+ \sum_{k=0}^{n-1} ( \, G(n,k+1) -  G(n,k) \, )
$$
$$
=\, F(n+1,n) \, +  \, G(n,n) - G(n,0) \, = \, (1-x^n)\cdots (1-x) + (- (1-x^n)\cdots (1-x)) - (-1)  \, = \, 1  \quad .
$$
Since $a(0)=0$, we have $a(n)=n$. The second identity is obtained from the first by replacing $x$ by $1/x$ and simplifying.

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