Since $p \mid n + 1$, then $p \mid n + 1 - p$. So, there exists $k \in \mathbb{N}$ such that $n = k p + p - 1$ and $\left\lfloor \frac{n}{p} \right\rfloor = k$. Now, we have
$$\begin{aligned} \binom{n}{p} - \left\lfloor \frac{n}{p} \right\rfloor &= \binom{k p + p - 1}{p} - k \\ &= \frac{(k p + p - 1)(k p + p - 2) \dots (k p + 1)(k p)}{p!} - k \\ &= \frac{k(k p + 1)(k p + 2) \dots (k p + p - 1) - k(p - 1)!}{(p - 1)!} \end{aligned}$$

$$= \frac{k(k \cdot p \cdot r + (p-1)! ) - k(p-1)!}{(p-1)!} = \frac{k^2 \cdot p \cdot r}{(p-1)!} \in \mathbb{N}$$
Since $\left(\left[\frac{n}{p}\right], (p-1)!\right) = 1$, then $(p-1)!$ divides $r$ and therefore $\binom{n}{p} - \left[\frac{n}{p}\right]$ is divisible by $p \cdot \left[\frac{n}{p}\right]^2$ as we wanted to prove. $\square$