Problem: Let  $k$ and $n$ be positive integers.
Prove that there exist positive integers $m_1$, \dots, $m_k$
such that
\[ 1 + \frac{2^k-1}{n} = \left( 1 + \frac{1}{m_1} \right) \left( 1 + \frac{1}{m_2} \right)
  \dots \left( 1 + \frac{1}{m_k} \right). \]

Solution: By induction on $k \ge 1$.
When $k = 1$ there is nothing to prove.

For the inductive step, if $n$ is even, write
\[
  \frac{n + (2^k-1)}{n}
  = \left( 1 + \frac{1}{n + (2^k-2)} \right) \cdot \frac{\frac n2 + (2^{k-1}-1)}{\frac n2}
\]
and use inductive hypothesis on the second term.
On the other hand if $n$ is odd then write
\[
  \frac{n + (2^k-1)}{n}
  = \left( 1 + \frac{1}{n} \right) \cdot \frac{\frac{n+1}{2} + (2^{k-1}-1)}{\frac{n+1}2}
\]
and use inductive hypothesis on the second term.