= Solution
Subtract the two finite sums. In the shifted sum replace the summation index $k$ by $k-1$. For every interior index the new <coefficient> is
$$
(-1)^k\binom rk-(-1)^{k-1}\binom r{k-1}=(-1)^k\binom{r+1}k
$$
by <Pascal's identity>; the two endpoint <coefficients> agree as well. Therefore
$$
\boxed{Q_{(r)}(x)-Q_{(r)}(x-h)=Q_{(r+1)}(x).}
$$
For a degree-$d$ <polynomial> with <leading coefficient> $c\neq0$, the backward difference has leading term
$$
c\bigl(x^d-(x-h)^d\bigr)=cdh\,x^{d-1}+\text{lower-degree terms}.
$$
Since $h>0$, this <coefficient> is nonzero for $d\geq1$. Starting with $Q_{(0)}=P$, iteration gives <degree> $n-r$ and <leading coefficient> $c h^r n!/(n-r)!$ for $0\leq r\leq n$. The next difference of the resulting constant is zero:
$$
\boxed{\deg Q_{(r)}=n-r\ (0\leq r\leq n),\qquad Q_{(n+1)}\equiv0.}
$$
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