How prove this binomial this problem  from book,he say it is clear have$$\sum_{k=0}^n\sum_{j=0}^{n-k}\binom{n}{n-k}\binom{n-k}{j}(a+tk)^{n-j-1}(-1)^{n-k-j}(a+b)^j
=a^{-1}(a+b)^n$$ where $a.b$ be real numbers,
why it is clear? if not,how to prove it?

$$
\begin{align}
&\sum_{k=0}^n\sum_{j=0}^{n-k}\binom{n}{n-k}\binom{n-k}{j}(a+tk)^{n-j-1}(-1)^{n-k-j}(a+b)^j\tag1\\
&=\sum_{k=0}^n\sum_{j=0}^n\binom{n}{n-k}\binom{n-k}{j}(a+tk)^{n-j-1}(-1)^{n-k-j}(a+b)^j\tag2\\
&=\sum_{j=0}^n\sum_{k=0}^n\binom{n}{n-j}\binom{n-j}{k}(a+tk)^{n-j-1}(-1)^{n-k-j}(a+b)^j\tag3\\
&=\underbrace{a^{-1}(a+b)^n\vphantom{\sum_{j=0}^{n-1}}}_{j=n}+\sum_{j=0}^{n-1}\binom{n}{n-j}\underbrace{\sum_{k=0}^{n-j}\binom{n-j}{k}\overbrace{(a+tk)^{n-j-1}}^{\substack{\text{degree $n-j-1$}\\\text{polynomial in $k$}}}(-1)^{n-k-j}}_{\text{$\Delta^{n-j}$ of a degree $n-j-1$ polynomial $=\ 0$}}(a+b)^j\tag4\\[6pt]
&=\bbox[5px,border:2px solid #C0A000]{a^{-1}(a+b)^n}\tag5
\end{align}
$$
Explanation:
$(2)$: the sum in $j$ can be extended to $n$ since $\binom{n-k}{j}=0$ for $j\gt n-k$
$(3)$: $\binom{n}{n-k}\binom{n-k}{j}=\binom{n}{n-j}\binom{n-j}{k}$ (write them out in terms of factorials)
$\phantom{(3)\text{: }}$and switch order of summation
$(4)$: pull the $j=n$ term out in front of the sum
$\phantom{(4)\text{: }}$when $j=n$, $\binom{n-j}{k}=0$ if $k\ne0$ so the sum in $k$ is trivial
$(5)$: an $n-j$ order repeated difference of a degree $n-j-1$ polynomial is $0$
