We prove the statement by induction on $m$. When $m = k$, we have
$$L_i \left[ \binom{k}{i} - \binom{0}{i} \right] = L_i \binom{k}{i} = L_i \cdot \frac{k!}{i!(k-i)!} = k \cdot \frac{L_i}{i} \cdot \frac{(k-1)!}{(i-1)!(k-i)!} = k \cdot \frac{L_i}{i} \cdot \binom{k-1}{i-1}.$$
Since $\frac{L_i}{i}$ is an integer (by definition of $L_i$), and $\binom{k-1}{i-1}$ is a binomial coefficient and thus also an integer, we see that $k$ is indeed a divisor.

For the induction step, assume that the statement holds for a specific value of $m$. We use the recursion $\binom{m+1}{i} = \binom{m}{i} + \binom{m}{i-1}$ to show that it holds for $m+1$ as well:
$$\begin{aligned} L_i \left[ \binom{m+1}{i} - \binom{m+1-k}{i} \right] &= L_i \left[ \binom{m}{i} - \binom{m-k}{i} \right] + L_i \left[ \binom{m}{i-1} - \binom{m-k}{i-1} \right] \\ &= L_i \left[ \binom{m}{i} - \binom{m-k}{i} \right] + \frac{L_i}{L_{i-1}} \cdot L_{i-1} \left[ \binom{m}{i-1} - \binom{m-k}{i-1} \right]. \end{aligned}$$
Note that $k$ divides both $L_i[\binom{m}{i} - \binom{m-k}{i}]$ and $L_{i-1}[\binom{m}{i-1} - \binom{m-k}{i-1}]$ by the induction hypothesis (if $i=1$, the latter term is simply zero), and $L_{i-1}$ (the least common multiple of $1, 2, \dots, i-1$) divides $L_i$ (the least common multiple of $1, 2, \dots, i$, or equivalently the least common multiple of $L_{i-1}$ and $i$). It follows that $k$ is also a divisor of $L_i[\binom{m+1}{i} - \binom{m+1-k}{i}]$, which completes the proof.