The [$2$-adic valuation](https://en.wikipedia.org/wiki/P-adic_order) of $n\in\mathbb{N}$, denoted $\nu(n)$, is the largest power $t$ such that $2^t$ divides $n$. The number of [integer partitions](https://en.wikipedia.org/wiki/Partition_(number_theory)) of $n$, denoted by $p(n)$, has generating function 
$$\sum\_{n\geq0}p(n)x^n=\prod\_{k\geq1}\frac1{1-x^k}.$$
However, I am finding an alternative:
$$\sum\_{n\geq0}p(n)x^n=\prod\_{k\geq1}(1+x^k)^{\nu(2k)}.$$



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> **Question.** Is this known? If so, any reference? If not, then any proof?
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**Caveat.** I would be surprised if this is new. 



is this a familiar gen. fn. for partitions?
This can be proved from the famous:



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> Distinct parts <-> Odd parts
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which can be found in Hardy & Wright : An Introduction to the Theory of Numbers.


This states:


$$(1+x)(1+x^2)(1+x^3)\dots=\frac1{(1-x)(1-x^3)(1-x^5)\dots}$$


If you substitute $x\to x^{2^k}$ for $k=1,\dots$, and multiply together, the RHS becomes the usual partition generating function, and the LHS takes your alternative form.


