Let's determine the greatest common divisor of the numbers $\binom{n}{1},\binom{n}{2},\binom{n}{3}, \ldots,\binom{n}{n-1}$.
Let the prime factorization of $n$ be $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{t}^{\alpha_{t}}$. For any $k \leq \alpha_{j}$ and $1 \leq j \leq t$,

$$
\binom{n}{p_{j}^{k}}=\frac{n}{p_{j}^{k}} \cdot \frac{n-1}{1} \cdot \frac{n-2}{2} \cdots \cdots \frac{n-\left(p_{j}^{k}-1\right)}{p_{j}^{k}-1}
$$

We will show that $\binom{n}{p_{j}^{k}}$ is not divisible by $p_{j}^{\alpha_{j}+1-k}$. Suppose that the numerator of one of the $\frac{n-i}{i}$ factors is divisible by $p_{j}^{v}$. The value of $v$ cannot be $\alpha_{j}$ or greater, because then $p_{j}^{\alpha_{j}}$ would divide both $(n-i)$ and $n$, and thus $i$, which is impossible given $1 \leq i \leq p_{j}^{k}-1 \leq p_{j}^{\alpha_{j}}-1$. Therefore, $p_{j}^{v}$ divides $n$ and thus $i=n-(n-i)$, the denominator of the fraction. Hence, if we simplify the fraction $\frac{n-1}{1} \cdot \frac{n-2}{2} \cdots \cdots \frac{n-\left(p_{j}^{k}-1\right)}{p_{j}^{k}-1}$, the numerator of the resulting fraction will not be divisible by $p_{j}$. Since $\frac{n}{p_{j}^{k}}$ is not divisible by $p_{j}^{\alpha_{j}+1-k}$, we have proven our statement. Specifically, we get that $\binom{n}{p_{j}^{\alpha_{j}}}$ is not divisible by $p_{j}$.

Let $m$ be the greatest common divisor of the numbers $\binom{n}{1}, \ldots,\binom{n}{n-1}$. From the above, it is clear that if $n$ is not a prime power, then $m$ is not divisible by any of the primes $p_{1}, p_{2}, \ldots, p_{t}$. Since $m$ is a divisor of $\binom{n}{1}=n$, the value of $m$ is 1.

Now let $n=p^{\alpha}$ be a prime power; clearly, in this case, $m=p^{\beta} \leq p^{\alpha}$ is possible only. Applying the previous observation for $k=\alpha-1$, we get that $\beta<\alpha+1-(\alpha-1)=2$, i.e., $\beta \leq 1$. On the other hand, for any $1 \leq l \leq p^{\alpha}-1$, we have $l=p^{w} \cdot l_{1}$, where $l_{1}$ is not divisible by $p$ and $w \leq \alpha-1$. Thus,

$$
\binom{n}{l}=\binom{p^{\alpha}}{l}=p \cdot p^{\alpha-1-w} \cdot \frac{p^{\alpha}-1}{l-1}
$$

Since $l_{1}$ is coprime to $p$, the quotient $\frac{\frac{p^{\alpha}-1}{l-1}}{l_{1}}$ is an integer, so $\binom{n}{l}$ is divisible by $p$. Therefore: $m=1$ if $n$ is not a prime power, and $m=p$ if $n$ is a power of the prime $p$.