$1000_2 - 0111_2$ = $8_{10}-7_{10} = 1_{10}$.
When there are no more on-bits to borrow from the number that is being subtracted from, is the resultings bits in the sum all zeros? (Se pic)
We know the result is $1$. So when there are no bits to borrow is the result zero?
Starting from the question,
$$\begin{array}{rrrrr} &1&0&0&0\\ -&&1&1&1\\ \hline \end{array}$$
From the rightmost digit, it needs to borrow from the left, so column-wise,
$$\begin{array}{rrrrr} &1&0&0&0\\ &&&\tiny{-1}&\tiny{+10}\\ -&&1&1&1\\ \hline\\ \end{array} \longrightarrow \begin{array}{rrrrr} &1&0&0&0\\ &&&\tiny{-1}&\tiny{+10}\\ -&&1&1&1\\ \hline &&&&1 \end{array}$$
Then the twos-digit, it also needs to borrow enough from the left to subtract 1 twice,
$$\begin{array}{rrrrr} &1&0&0&0\\ &&&\tiny{-1}&\tiny{+10}\\ &&\tiny{-1}&\tiny{+10}\\ -&&1&1&1\\ \hline &&&&1 \end{array} \longrightarrow \begin{array}{rrrrr} &1&0&0&0\\ &&&\tiny{-1}&\tiny{+10}\\ &&\tiny{-1}&\tiny{+10}\\ -&&1&1&1\\ \hline &&&0&1 \end{array}$$
Then the fours-digit and up,
$$\begin{array}{rrrrr} &1&0&0&0\\ &&&\tiny{-1}&\tiny{+10}\\ &&\tiny{-1}&\tiny{+10}\\ &\tiny{-1}&\tiny{+10}\\ -&&1&1&1\\ \hline &&&0&1 \end{array} \longrightarrow \begin{array}{rrrrr} &1&0&0&0\\ &&&\tiny{-1}&\tiny{+10}\\ &&\tiny{-1}&\tiny{+10}\\ &\tiny{-1}&\tiny{+10}\\ -&&1&1&1\\ \hline &0&0&0&1 \end{array}$$
So $$1000-111=1$$