Expanding $(1+x/j)^{-1}(1+1/j)^{x}$ In introducing the gamma function in chapter 1 of the Special Functions by Pugh et al it is written without proof

$(1+\frac{x}{j})^{-1}(1+\frac{1}{j})^{x}=1 + \frac{x(x-1)}{2j^2 } + O (\frac{1}{j^3} ) $ for $x \in \mathbb{C} - {\{-1,-2,...}\}$ and $j \in \mathbb{N}.$ and therefore $\Pi_{j=1}^{\infty}(1+\frac{x}{j})^{-1}(1+\frac{1} {j})^{x}$ converges. 

For me even with intermediate knowledge of real and complex analysis I have , this is too much to handle rigorously!
Questions :
1-How this expansion happens with attention to the given domains of x and j?
2- how the product converges?
Edit. I could solve the question 2 rigorously so only the question 1 remains unsolved. Thank you 
 A: You are using Euler's definition of the Gamma function to write 
$$  x! = \prod_{j=1}^\infty \left( \left( 1+\frac{x}{j} \right)^{-1} \left( 1+\frac{1}{j} \right)^x \right)  \text{,}  $$
so we only care about the case that $j$ is a positive integer and $j$ will range through all positive integers.
In the term $\left( 1 + \frac{x}{j} \right)^{-1}$, we cannot have $x = -j$, otherwise, we introduce division by zero, so $x \not\in\{-1,-2,-3,\dots\}$.  As there are no other opportunities for the expression 
$$  \left( 1+\frac{x}{j} \right)^{-1} \left( 1+\frac{1}{j} \right)^x  $$ to be undefined, we attempt expansion in Laurent series.
First,
$$  \left( 1+\frac{x}{j} \right)^{-1} = 1 - \frac{x}{j} + \frac{x^2}{j^2} - \cdots = \sum_{k=0}^\infty \left( \frac{-x}{j} \right) ^k  \text{,}  $$
which is a convergent geometric series for $|x| < |j|$.  For any particular choice of $x$, we break the product into the finite initial segment where we do not use the series, $\prod_{j=1}^{\lfloor |x| \rfloor}$, and the following infinite segment where the series converges to the term.  Questions of convergence rest on analysis of the second product.
Second,
$$  \left( 1 + \frac{1}{j}\right)^x = 1 + \frac{x}{j} + \frac{x(x-1)}{2 j^2} + \cdots = \sum_{k=0}^\infty \frac{x^\underline{k}}{k! j^k}  \text{,}  $$
where we have used $x^{\underline{k}}$ to represent the falling factorial, 
$$  x^{\underline{k}} = x(x-1)\cdots (x-k+1)    $$
with the usual convention for empty products: $x^{\underline{0}} = 1$.  This series is dominated by that of $\mathrm{e}^{x/j}$, so represents an entire function.
Keeping the terms shown, multiplying out and again discarding terms with power of $j$ less than $-2$ (tracked by our use of big-O notation), we have
$$  \left( 1+\frac{x}{j} \right)^{-1} \left( 1+\frac{1}{j} \right)^x = 1 + \frac{x(x-1)}{2j^2} + O(j^{-3})  \text{.}  $$
So now we consider 
$$  A = \prod_{j = \lceil |x| \rceil}^{\infty} \left( 1 + \frac{x(x-1)}{2j^2} + O(j^{-3}) \right) $$
and, as is common for studying convergence of products, we consider the convergence of the equivalent series
$$  \mathrm{e}^{\ln A} = \exp \left( \sum_{j = \lceil |x| \rceil}^{\infty} \ln \left( 1 + \frac{x(x-1)}{2j^2} + O(j^{-3}) \right) \right)  $$
Using the familiar series for the logarithm centered at $1$ (and noting that $j > |x|$, so this series converges for the argument we are using) and that the logarithm is concave down, 
\begin{align*}
\left| \mathrm{e}^{\ln A} \right| &\leq \left| \exp \left( \sum_{j=\lceil |x| \rceil}^{\infty} \frac{x(x-1)}{2j^2} + O(j^{-3}) \right) \right|  \end{align*}
The integral test using $\int^\infty j^{-2} \,\mathrm{d}j$ (and, if needed, $\int^\infty K j^{-3} \,\mathrm{d}j$, where $K$ is the bounding constant hiding in the big-O) show that the series converges, so the product converges.
A: Hints:


*

*Calculate the Taylor expansions of $f_1(t)=(1+xt)^{-1}$ and $f_2(t)=(1+t)^x$ at $t=0$, in the form $f_j(t) = f_j(0) + f_j'(0)t + \frac12f_j''(0)t^2 + O(|t|^3)$, where the constant in the error term is allowed to depend on $f_j$ and on $x$. The first Taylor expansion will hold for $|t|<1/|x|$, the second for $|t|<1$.

*Then plug in $t=1/j$ to both Taylor expansions. For the first function, this might only hold if $j$ is large enough in terms of $|x|$; can you argue that if the asymptotic expansion holds for large $j$ then it holds for all $j$, possibly with a larger implicit constant?

*Convince yourself that any product of the form $\prod_{j=1}^\infty \big( 1+O(\frac1{j^2}) \big)$ converges (hint: take logarithms for $j$ sufficiently large).

