Hermite's First Divisibility Property for Binomial Coefficient $\binom{n}{r}$ states, for $n,r \geq 1$,
$\frac{n}{gcd(n,r)} | \binom{n}{r}$ .
Obviously, $\frac{n}{gcd(n,r)}$ is a divisor of $n$.
But is $\frac{n}{gcd(n,r)}$ the largest divisor of $n$ that divides $\binom{n}{r}$?
Claim: $m>n>0$, $p$ is a prime number, $\alpha >0$ such that $p^\alpha || m$ (that is, $p^\alpha \mid m$ and $p^{\alpha+1} \nmid m$). If $p^\beta || C(m, n)$, then $\beta \ge \alpha - s$, where $s\ge 0$ and $p^s || n$.
Proof of claim: $$\beta =\sum_{i=1}^{\infty} ([\frac{m}{p^i}] - [\frac{n}{p^i}] - [\frac{m-n}{p^i}])$$
Each term in the summation contributes $0$ or $1$. If $n=1$, the first $\alpha$ terms contribute $1$ each; for general $n$, except the first $s$ terms, the contribution of each term is at least as good as in $C(m, 1)$.