= Solution
Put $A=\operatorname{im}\alpha$. The restriction $\beta|_A$ has image $\operatorname{im}(\beta\alpha)$ and kernel $A\cap\ker\beta$. By the <rank-nullity theorem>,
$$
r(\beta\alpha)=r(\alpha)-\dim(A\cap\ker\beta).
$$
Since its image is contained in $\operatorname{im}\beta$, this proves the upper bound. Also $\dim(A\cap\ker\beta)\le\dim\ker\beta=\dim V-r(\beta)$, giving the <rank inequality for a composition>:
$$
\boxed{r(\alpha)+r(\beta)-\dim V\le r(\beta\alpha)\le\min(r(\alpha),r(\beta)).}
$$
Examples on $U=V=W=\mathbb R^2$ distinguish the inequalities. Taking $\alpha=\beta=I$ gives equality in both. Taking $\alpha=\operatorname{diag}(1,0)$ and $\beta=\operatorname{diag}(0,1)$ gives composition rank zero, strictly below the upper bound 1 and equal to the lower bound 0. Taking $\alpha=\beta=\operatorname{diag}(1,0)$ gives rank 1, strictly above the lower bound 0 and equal to the upper bound 1.
For the final <nilpotent linear map>, repeated application of the lower bound gives $r(\alpha^k)\ge2n-2k$. Since $\alpha^n=\alpha^{n-k}\alpha^k=0$, another application gives
$$
0\ge r(\alpha^{n-k})+r(\alpha^k)-2n
\ge2k+r(\alpha^k)-2n.
$$
The two bounds coincide, yielding
$$
\boxed{r(\alpha^k)=2n-2k\quad(1\le k\le n-1).}
$$
In fact the same formula includes $k=0,n$. For $n=1$ the requested index range is empty and the assumptions force $\alpha=0$.
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