$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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The first OP statement is given by the enclosed expression at the very beginning:
\begin{align}
&\bbx{\sum_{x_{1} = \color{#f00}{\large 1}}^{\infty}
\sum_{x_{2} = \color{#f00}{\large 1}}^{\infty}
\sum_{x_{3} = \color{#f00}{\large 1}}^{\infty}\cdots
\sum_{x_{n} = \color{#f00}{\large 1}}^{\infty}
\bracks{z^{a}}z^{\large \,x_{1} + 2x_{2} + 3x_{3} + \cdots + nx_{n}}}
\\[5mm] = &\
\sum_{x_{1} = \color{#f00}{\large 0}}^{\infty}
\sum_{x_{2} = \color{#f00}{\large 0}}^{\infty}
\sum_{x_{3} = \color{#f00}{\large 0}}^{\infty}\cdots
\sum_{x_{n} = \color{#f00}{\large 0}}^{\infty}\bracks{z^{a}}
z^{\large \,\pars{x_{1} + 1} + 2\pars{x_{2} + 1} + 3\pars{x_{3} + 1} + \cdots + n\pars{x_{n} + 1}}
\\[5mm] = &\
\sum_{x_{1} = \color{#f00}{\large 0}}^{\infty}
\sum_{x_{2} = \color{#f00}{\large 0}}^{\infty}
\sum_{x_{3} = \color{#f00}{\large 0}}^{\infty}\cdots
\sum_{x_{n} = \color{#f00}{\large 0}}^{\infty}
\bracks{z^{a}}z^{\large n\pars{n + 1}/2\ +\ \,x_{1} + 2x_{2} + 3x_{3} + \cdots + nx_{n}}
\\[5mm] = &\
\bbx{\sum_{y_{1} = \color{#f00}{\large 0}}^{\infty}
\sum_{y_{2} = \color{#f00}{\large 0}}^{\infty}
\sum_{y_{3} = \color{#f00}{\large 0}}^{\infty}\cdots
\sum_{y_{n} = \color{#f00}{\large 0}}^{\infty}\bracks{z^{a - n\pars{n + 1}/2}}
z^{\large \,y_{1} + 2y_{2} + 3y_{3} + \cdots + ny_{n}}}
\end{align}
which is the OP second statement.