I assume that $p$ is prime.
Kummer's Theorem says that the number of factors of $p$ in $\binom{n}{k}$ is the sum of the base-$p$ digits of $n$ minus the sum of the base-$p$ digits of $k$ and $n-k$, all divided by $p-1$.
In the case above, the base-$p$ representation of $n$ looks like $ d_md_{m-1}d_{m-2}\dots d_{\alpha}\underbrace{000\dots000}_{\alpha\text{ zeros}} $ where $d_\alpha\not=0$. Furthermore, $p^\alpha$ looks like $ 1\underbrace{000\dots000}_{\alpha\text{ zeros}} $ and $n-p^\alpha$ looks like $ d_md_{m-1}d_{m-2}\dots (d_{\alpha}-1)\underbrace{000\dots000}_{\alpha\text{ zeros}} $ The sum of the base-$p$ digits in $n-p^\alpha$ and $p^\alpha$ is $ \sum_{j=\alpha}^md_j $ which is the same as the sum of the base-$p$ digits in $n$. Thus, there are no factors of $p$ in $\binom{n}{p^\alpha}$.