is this a familiar gen. fn. for partitions? The $2$-adic valuation of $n\in\mathbb{N}$, denoted $\nu(n)$, is the largest power $t$ such that $2^t$ divides $n$. The number of integer partitions 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:
$$\sum_{n\geq0}p(n)x^n=\prod_{k\geq1}(1+x^k)^{\nu(2k)}.$$

Question. Is this known? If so, any reference? If not, then any proof?

Caveat. I would be surprised if this is new. 
 A: This can be proved from the famous:

Distinct parts <-> Odd parts

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.
A: I'm going to give this another go because the method reveals some points of potential interest. 
The intent is to prove that $\prod_{k\geq1}\frac1{1-x^k}=\prod_{k\geq1}(1+x^k)^{\nu(2k)}$. Take logarithms on both sides and make use the Taylor series $-\log(1-y)=\sum_{i\geq1}\frac{y^i}i$. The LHS turns into
\begin{align} \log\prod_{k\geq1}\frac1{1-x^k}=-\sum_{k\geq1}\log(1-x^k)=\sum_{m,k\geq1}\frac{x^{mk}}k
=\sum_{n\geq1}x^n\sum_{d\vert n}\frac1d=\sum_{n\geq1}\frac{x^n}n\sigma(n),
\end{align}
where $\sigma(n)$ is the sum of divisors function.
From $\log(1+y)=-\sum_{i\geq1}(-1)^i\frac{y^i}i$, the RHS converts to
\begin{align} \log\prod_{k\geq1}(1+x^k)^{\nu(2k)}
&=\sum_{k\geq1}\nu(2k)\cdot\log(1+x^k)
=-\sum_{m,k\geq1}\frac{(-1)^mx^{mk}\nu(2k)}m \\
&=-\sum_{n\geq1}x^n\sum_{d\vert n}(-1)^{n/d}\frac{\nu(2d)}{n/d}
=-\sum_{n\geq1}\frac{x^n}n\sum_{d\vert n}(-1)^{n/d}d\cdot\nu(2d).
\end{align}
In other words, we need to prove
$$\sigma(n)=-\sum_{d\vert n}(-1)^{n/d}d\cdot\nu(2d).$$
Consider the Dirichlet series $f(s):=-\sum_{a\geq1}\frac{(-1)^a}{a^s}$ and $g(s):=\sum_{b\geq1}\frac{\nu(b)}{b^s}$ to obtain
\begin{align} f(s+1)g(s)
&=-\sum_{a\geq1}\frac{(-1)^a}{b^{s+1}}\sum_{b\geq1}\frac{b\cdot\nu(b)}{b^{s+1}}
=-\sum_{n\geq1}\frac1{n^{s+1}}\sum_{d\vert n}(-1)^{n/d}d\cdot\nu(d). \tag1
\end{align}
It's easy to see $f(s+1)=(1-2^{-s})\zeta(s+1), g(s)=\frac{\zeta(s)}{2^s-1}$ and $\zeta(s+1)\zeta(s)=\sum_{n\geq1}\frac{\sigma(n)}{n^{s+1}}$ where $\zeta(s)$ is the Riemann zeta function. Therefore, we have
$$f(s+1)g(s)=\frac1{2^s}\sum_{n\geq1}\frac{\sigma(n)}{n^{s+1}}
=\sum_{n\geq1}\frac{2\sigma(n)}{(2n)^{s+1}}.\tag2$$
Reading off the coefficients of $\frac1{(2n)^{s+1}}$, from the Dirichlet generating functions in (1) and (2), we get
$$2\sigma(n)=-\sum_{d\vert 2n}(-1)^{2n/d}d\cdot\nu(d)
=-\sum_{2d'\vert 2n}(-1)^{2n/(2d')}2d'\cdot\nu(2d') \tag3$$
up on using $\nu(odd)=0$. Rewrite equation (3) to obtain the desired result:
$$\sigma(n)=-\sum_{d'\vert n}(-1)^{n/d'}d'\cdot\nu(2d').$$
A: We have the identity
$$\frac{1}{1 - x^k} = \prod_{i \ge 0} (1 + x^{k \cdot 2^i})$$
which is equivalent to the uniqueness of binary representations, and is also straightforward to prove using a telescoping argument by multiplying both sides by $1 - x^k$. Applying this identity to every term on the RHS of the first identity produces
$$\sum_{n \ge 0} p(n) x^n = \prod_{k \ge 1} \prod_{i \ge 0} (1 + x^{k \cdot 2^i})$$
which rearranges to your identity without much effort. The combinatorial interpretation here involves looking at the binary representation of the multiplicities of each integer in a partition. 
