Proof: When $l=1,2$, the conclusion can be directly verified. Now assume $l \geqslant 3$. Similarly, by property VIII of §1 and the subsequent explanation, for each $r_{i}$, there must be a unique $r_{j}$ such that
$$r_{i} r_{j} \equiv 1\left(\bmod 2^{l}\right)$$
The necessary and sufficient condition for $r_{i}=r_{j}$ is
$$r_{i}^{2} \equiv 1\left(\bmod 2^{l}\right),$$

which is equivalent to
$$\left(r_{i}-1\right)\left(r_{i}+1\right) \equiv 0\left(\bmod 2^{l}\right)$$

Noting that $\left(r_{i}, 2\right)=1$, the above equation becomes
$$\frac{r_{i}-1}{2} \cdot \frac{r_{i}+1}{2} \equiv 0\left(\bmod 2^{l-2}\right)$$

Noting that
$$\left(\frac{r_{i}-1}{2}, \frac{r_{i}+1}{2}\right)=1$$

it follows that the necessary and sufficient condition for $r_{i}=r_{j}$ is
$$\frac{r_{i}-1}{2} \equiv 0\left(\bmod 2^{l-2}\right) \quad \text { or } \quad \frac{r_{i}+1}{2} \equiv 0\left(\bmod 2^{l-2}\right),$$

which is equivalent to
$$r_{i} \equiv 1\left(\bmod 2^{l-1}\right) \quad \text { or } \quad r_{i} \equiv-1\left(\bmod 2^{l-1}\right).$$

Therefore, in the reduced residue system modulo $2^{l}$, $r_{i}=r_{j}$ only when
$$r_{i} \equiv 1,2^{l-1}+1,2^{l-1}-1 \text { or } 2^{l}-1\left(\bmod 2^{l}\right).$$

Thus, for each $r_{i}$ in the reduced residue system modulo $2^{l}$, except for these four numbers (which are pairwise incongruent modulo $2^{l}$), there must be $r_{j} \neq r_{i}$. Therefore, except for these four numbers, the $c-4$ numbers in the reduced residue system can be paired off according to equation (9), meaning the product of these $c-4$ numbers is congruent to 1 modulo $2^{l}$. This, together with equation (10), proves that equation (8) holds for $l \geqslant 3$.