\[\boxed{\mathbf{887\ (887).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[a^{n} - 1 = (a - 1)\]
\[(a^{n - 1} + a^{n - 2} + \ldots + a + 1)\]
\[\left( a^{n - 1} + a^{n - 2} + \ldots + a + 1 \right) =\]
\[= S_{n} = \frac{b_{1}\left( q^{4} - 1 \right)}{q - 1} -\]
\[геометрическая\ прогрессия.\]
\[b_{1} = 1,\ \ q = \frac{a}{1} = a\ \]
\[S_{n} = \frac{1 \cdot \left( a^{n} - 1 \right)}{a - 1} = \frac{a^{n} - 1}{a - 1}\]
\[a^{n} - 1 = (a - 1) \cdot \frac{a^{n} - 1}{a - 1}\]
\[\ \frac{a^{n} - 1}{a - 1} = \frac{a^{n} - 1}{a - 1} \Longrightarrow доказано.\]