<s>[INST] Decomposition as a product of factors For several days, I try to do this excercise. Without success.
Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1, m_2, \dots, m_k$ (not necessarily different) such that:
$$1+ \frac{2^k - 1}n = \left(1+\frac 1{m_1}\right)\cdot\left(1+\frac 1{m_2}\right)\cdots\left(1+\frac 1{m_k}\right)$$
 [/INST] $k=1$ is clear.  We now suppose that the statement is true for some fixed $k$ (and all $n$), and proceed to demonstrate it for $k+1$.
If $n$ is even, then we can write $1+\frac{2^k-1}{\frac{n}{2}}$ as a product of the given form with $k$ terms.  And so, the following can be written as a product with $k+1$ terms:
$$\left(1+\frac{1}{n+2^{k+1}-2}\right)\left(1+\frac{2^k-1}{\frac{n}{2}}\right) = \left(\frac{n+2^{k+1}-1}{n+2^{k+1}-2}\right)\left(\frac{n+2^{k+1}-2}{n}\right)$$
$$= \frac{n+2^{k+1}-1}{n} = 1 + \frac{2^{k+1}-1}{n}$$
This verifies the claim when $n$ is even.  If $n$ is odd, then we can write $1+\frac{2^k-1}{\frac{n+1}{2}}$ as a product with $k$ terms.  As above, the following is a product with $k+1$ terms:
$$\left(1+\frac{1}{n}\right)\left(1+\frac{2^k-1}{\frac{n+1}{2}}\right) = \left( \frac{n+1}{n}\right)\left( \frac{n+2^{k+1}-1}{n+1}\right)$$
$$= \frac{n+2^{k+1}-1}{n} = 1 + \frac{2^{k+1}-1}{n}$$
That about does it.
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