A conjecture regarding products of $u(x)=x+\frac1x$ Recently, in a pre-calculus textbook I saw lying around my high school, I saw a problem having something to do with the neat identity 
$$\left(a+\frac1a\right)\left(b+\frac1b\right)\left(ab+\frac1{ab}\right)=2+a^2+\frac1{a^2}+b^2+\frac1{b^2}+a^2b^2+\frac1{a^2b^2}.$$
The details of the problem aren't too important, but it got me thinking that there must be a generalization. To make more apparent the cool nature of this identity I rephrase it in terms of the function $u(x)=x+\frac1{x}$:
$$u(a)u(b)u(ab)-u(1)=u(a^2)+u(b^2)+u(a^2b^2).$$
Indeed, there is an analog to this formula:
$$\begin{align}
u(a)u(b)u(c)u(abc)-u(1) &= u(a^2)+u(b^2)+u(c^2)\\
&+u(a^2b^2)+u(a^2c^2)+u(b^2c^2)\\
&+u(a^2b^2c^2).
\end{align}$$
And yet another analog:
$$u^2(a)-u(1)=u(a^2).$$
This pattern was immediately recognizable to me, and I formed a conjecture. 
Fix any $n\in \Bbb N$.

Conjecture: For $a_k\ne 0$, $k=1,2,...,n$, we have
  $$\Pi_n:=u\left(\small{\prod_{k=1}^{n}}a_k\right)\prod_{k=1}^{n}u(a_k)=u(1)+\sum_{k=1}^{n}\sum_{\sigma\subset S_n\\ |\sigma|=k}u\left(\small{\prod_{j\in\sigma}}a_j^2\right),\tag{1}$$ where $S_n=\{1,2,...,n\}$.

As you can see, $(1)$ holds for $n=1,2,3$. The process is very horrible, but I was able to verify the $n=4$ case as well. I have not yet verified any $n\ge5$, as the algebra gets really messy really quickly. My hope is to bypass this messy process by attacking the general case directly. I plan to do this by using the fact that, given a finite set $A$ of size $n$, we have
$$\prod_{p\in A}(x+p)=\sum_{k=0}^{n}x^k\sum_{\sigma\subset A\\ |\sigma|=n-k}\prod_{p\in\sigma}p,$$ 
so that $$\Pi_n=u(P)\prod_{k=1}^{n}\frac{1+a_k^2}{a_k}=\frac{P^2+1}{P^2}\prod_{k=1}^{n}(1+a_k^2) \qquad [\text{where }P=\prod_{k=1}^{n}a_k]$$
is the same as
$$\Pi_n=\frac{P^2+1}{P^2}\sum_{k=0}^{n}\sum_{\sigma\subset S_n\\ |\sigma|=k}\prod_{j\in\sigma}a_j^2.$$
This looks close enough to give me hope for the future, but not close enough to see all the way to the end of the tunnel. 
Any hints on how to proceed? Thanks. 
Also, if you have any other proofs I would greatly appreciate it.
 A: The key fact to know is
$$ u(x)u(y) = (x+1/x)(y+1/y) = u(xy)+u(x/y). \tag{1} $$
Given  non-zero numbers $\,a_1,a_2,\dots\,$ 
define the following sequences:
$$ y_n := \prod_{k=1}^n a_n, \;\;
 P_n := \prod_{k=1}^n u(a_n), \;\;
Q_n(t) := P_n\, u(t^2y_n). \tag{2} $$
It is easy to check that
$$ Q_n(t) = Q_{n-1}(t) + Q_{n-1}(t\,x_n) \tag{3} $$
expands into a $\,2^n\,$ term sum with one term for each divisor of $\,y_n.\,$
Your conjecture is that
$$ \Pi_n = Q_n(1) \tag{4} $$
which is true because $Q_n(t)$ is defined as a product of $\,n+1\,$ factors in equation $(2)$ and
expands into a $\,2^n\,$ term sum via equation $(3)$.
A: Maybe we could break the product into two expression:
$$\Pi_n:=u\left(\prod_{K=1}^n a_k\right)\prod_{K=1}^n u(a_k)$$
$$=\left(\prod_{K=1}^n a_k\right)\left(\prod_{K=1}^n u(a_k)\right)+\left(\prod_{K=1}^n \frac{1}{a_k}\right)\left(\prod_{K=1}^n u(a_k)\right)$$
$$=\prod_{K=1}^n a_k\cdot u(a_k)+\prod_{K=1}^n \frac{1}{a_k}\cdot u(a_k)$$
$$=\prod_{K=1}^n (a_k^2+1)+\prod_{K=1}^n \left(\frac{1}{a_k^2}+1\right)$$
Now try to rewrite $\prod (a_k^2+1)$ and $\prod\left(\frac{1}{a_k^2}+1\right)$ as sums (you already did it for $\prod (a_k^2+1)$)
A: Hint:
Just translating your language a little bit, consider $A=\{a_1,\cdots ,a_n\}$ and consider a weight on a set $q(S)=\prod _{s\in S}a_s$ (denote $q(\emptyset)=1$, that is where the $2$ is coming.) Here $u:\mathcal{P}(S_n)\longrightarrow \mathbb{Z}[a_1,\cdots a_n,\frac{1}{a_1},\cdots a_n]$ is defined as $u(S)=q(S)+q(S)^{-1}.$ 
Show that $$\prod _{i=1}^n u(\{i\})=\sum _{\substack{A\cup B=S_n\\A\cap B=\emptyset}}\frac{q(A)}{q(B)}.$$ here an induction argument suffices.
Show that $q(X)=q(A)q(B)$ if $A\cup B=X$ with $A\cap B=\emptyset.$  Notice that $A,B$ gives you two summands in the RHS of the equation above.
