Question concerning trace over tensor products I am trying to prove a certain theorem, which requires me to understand a certain manipulation.
Let $\{E_n\}_{n\ge0}$ denote the eigenvalues, in increasing order, of an operator $H$ on a Hilbert space $\mathcal H.$ It follows that the trace $\mathrm{Tr}_{_{\mathcal H}}(e^{-H}) = \sum_{j_1}e^{-E_{j_1}} .$
Let $\mathcal H^m := \bigotimes_{p=1}^m \mathcal H$ and $\mathcal H^0:= \mathbb C$. I would like to compute the trace of $\bigotimes_{p=1}^m e^{-H}$ over $\mathcal H^m.$
$$ \mathrm{Tr}_{_{\mathcal H^m}}(\bigotimes_{p=1}^m e^{-H}) = \prod_{p=1}^m \ \sum_{\color{red}{0\le j_1\le...\le j_m}} e^{- E_{j_p}} $$
Here's my first question. Why is the ordering (highlighted in $\color{red}{\text{red}}$) above necessary?
Secondly, why are the following equality and the subsequent inequality true? 
$$\sum_{m \ge 0} \ \prod_{p=1}^m \ \sum_{0\le j_1\le...\le j_m} z^m e^{- E_{j_p}} = \prod_{m\ge 0} (1+ze^{-E_m}) \le \prod_{m\ge 0} exp(ze^{- E_m})$$
Thank you for your time.

EDIT: The sum over highlighted indices should be a strict ordering as follows:
$$ \mathrm{Tr}_{_{\mathcal H^m}}(\bigotimes_{p=1}^m e^{-H}) = \prod_{p=1}^m \ \sum_{\color{red}{0\le j_1\lt...\lt j_m}} e^{- E_{j_p}} $$
Please refer to Operator Algebras and Quantum Statistical Mechanics II Proposition 5.2.22 (pp.46-48) to verify this result.
 A: The equation $$\mathrm{Tr}_{_{\mathcal H^m}}(\bigotimes_{p=1}^m e^{-H}) = \prod_{p=1}^m \ \sum_{{0\le j_1<...< j_m}} e^{- E_{j_p}}$$ is incorrect.  As a toy example, let's take $m=2$ and $\mathcal{H}$ to be two-dimensional, so we only have two eigenvalues $E_0$ and $E_1$.  The eigenvalues of $e^{-H}\otimes e^{-H}$ are then $e^{-E_0}e^{-E_0}$, $e^{-E_0}e^{-E_1}$, $e^{-E_1}e^{-E_0}$, and $e^{-E_1}e^{-E_1}$.  But in your equation, ${0\leq j_1<...< j_m}$ would give only one term: $0<1$.  This would then say that the trace is $e^{-E_0}\cdot e^{-E_1}$.  This is obviously incorrect.
The correct expression for the trace is just $$\prod_{p=1}^m\sum_{n\geq 0} e^{-E_n},$$ since the trace of a tensor product of operators is just the product of the traces of the individual operators.
Your expression for the trace would be almost correct if $\mathcal{H}^m$ instead referred to the $m$th exterior power of $\mathcal{H}$, which appears to be what the book you are referring to is talking about.  But you have the sum and product reversed: the correct expression for the trace would then be $$\sum_{{0\le j_1<...< j_m}}\prod_{p=1}^m e^{- E_{j_p}}.$$ The ordering $0\le j_1<...< j_m$ comes from the antisymmetrization of the exterior power: a basis for the exterior power consists of products of basis vectors for $\mathcal{H}$ in ascending order like this, since if you have two copies of the same basis vector that becomes $0$ in the exterior power, and if you have two basis vectors that are not in order you can just swap them (and get a minus sign).
The equation $$\sum_{m \ge 0} \ \prod_{p=1}^m \ \sum_{0\le j_1<...< j_m} z^m e^{- E_{j_p}} = \prod_{m\ge 0} (1+ze^{-E_m})$$ is not correct either.  What is true is 
$$\sum_{m \ge 0} z^m \sum_{0\le j_1<...< j_m} \prod_{p=1}^m e^{- E_{j_p}} = \sum_{m \ge 0} \ \sum_{0\le j_1<...< j_m}\prod_{p=1}^m z e^{- E_{j_p}} = \prod_{n\ge 0} (1+ze^{-E_n}).$$
Here the first equality comes from bringing $z^m$ inside the product and sum (note that since you have a product of $m$ factors, you get just one $z$ in each factor from $z^m$), and the second equality comes from just grouping terms of the expanded product $\prod_{n\ge 0} (1+ze^{-E_n})$ according to how many factors of the form $ze^{-E_n}$ they have, with $m$ being that number.  (Here I am glossing over convergence issues in order to assume you can ignore terms with infinitely many factors of $ze^{-E_n}$.)
Finally, the inequality $$\prod_{m\ge 0} (1+ze^{-E_m}) \le \prod_{m\ge 0} \exp(ze^{- E_m})$$ is just the fact that $1+w\leq \exp(w)$ for any $w\in\mathbb{R}$.  To prove this inequality, for instance, you could consider the function $f(w)=\exp(w)-w-1$ and note that $f'(w)<0$ for $w<0$, $f'(0)=0$, and $f'(w)>0$ for $w>0$, so $f$ has a global minimum at $w=0$.
