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Daca \( p \) este un numar natural prim si \( n\in \mathbb{N}^* \) sa se demonstreze ca
\( ({p^n\choose 1}, {p^n\choose 2},\ldots, {p^n\choose p^{n-1}})=p \).

Fie \( d = \gcd_{k \in \overline{1,p^{n-1}}} \left\{ {p^n \choose k} \right\} \). Atunci \( d | {p^n \choose 1} = p^n \), de unde \( d = p^t \), cu \( t \in \overline{1, n} \), \( t = \min_{k \in \overline{1,p^{n-1}}} \left\{ v_p\left( {p^n \choose k} \right) \right\} \).
Pe de o parte,
\( t \le v_p\left( {p^n \choose p^{n-1}} \right) = \sum_{j \ge 1} \left( \left[ \frac{p^n}{p^j} \right] - \left[ \frac{p^{n-1}}{p^j} \right] - \left[ \frac{p^n - p^{n - 1}}{p^j} \right] \right) = 1 \).
Pe de alta parte, pentru orice \( k \in \overline{1, p^{n-1}} \),
\( v_p\left( {p^n \choose k} \right) = \sum_{t \ge 1} \left( \left[ \frac{p^n}{p^j} \right] - \left[ \frac{k}{p^j} \right] - \left[ \frac{p^n - k}{p^j} \right] \right) \ge \left[ \frac{p^n}{p^n} \right] - \left[ \frac{k}{p^n} \right] - \left[ \frac{p^n - k}{p^n} \right] = 1 \),
de unde \( t \ge 1 \).
Din ceea ce am zis mai sus, \( t = 1 \), ceea ce trebuia aratat.