Let $m = k + t \times l \times (k!)$, where $t$ is any positive integer. To prove that $\binom{m}{k}$ and $l$ are relatively prime, we only need to prove that for any prime factor $p$ of $l$, $p \nmid \binom{m}{k}$.

If $p \nmid k!$, we have
$$\begin{aligned} k! \binom{m}{k} &= \prod_{i=1}^{k} (m-k+i) \\ &= \prod_{i=1}^{k} [i + t l (k!)] \\ &= \prod_{i=1}^{k} i \equiv k! \pmod{p}. \end{aligned}$$
Therefore, $p \nmid \binom{m}{k}$.

If $p \mid k!$, there exists integer $\alpha \ge 1$ such that $p^\alpha \mid k!$ but $p^{\alpha+1} \nmid k!$. Then $p^{\alpha+1} \mid l(k!)$.
We have
$$\begin{aligned} k! \binom{m}{k} &= \prod_{i=1}^{k} (m-k+i) \\ &= \prod_{i=1}^{k} [i + t l (k!)] \\ &= \prod_{i=1}^{k} i \equiv k! \pmod{p^{\alpha+1}}. \end{aligned}$$
Therefore, $p^\alpha \mid k! \binom{m}{k}$ and $p^{\alpha+1} \nmid k! \binom{m}{k}$. Since $p^\alpha \mid k!$, we get $p \nmid \binom{m}{k}$. The proof is completed.

Let $m = k + t \times l \times (k!)^z$, where $t$ is any positive integer. The following proof steps are similar to those in Solution I, and are omitted.