Show that $(x+1+O(x^{-1}))^x = ex^x + O(x^{x-1})$ for $x\rightarrow \infty$ So I'm trying to show that for $x\rightarrow \infty$:
$$(x+1+O(x^{-1}))^x = ex^x + O(x^{x-1})$$
So these complicated big-Oh expressions are clearly going to be a recurring theme in my book, and I simply have no idea how to manipulate them in a rigorous fashion.  I know logically what these expressions are saying, for instance this one means that:  
For any function $f(x)=O(x^{-1})$ for all $a<x< \infty$ for some $a$, we must prove that there exists some $b$ such that $(x+1+O(f(x)))^x = ex^x + O(x^{x-1})$ for all $b< x< \infty$.  And then I guess just use the $Max(a,b)$.
Nevertheless my book provides no list of 'legal' operations for big-Oh notation, and no examples of how to work with expressions that don't fall into the most simple category of $f(x) = O(g(x))$.  Thus I'm simply at a loss not only for how I go about proving such expressions but I don't even know what operations I'm allowed to perform.  How does one generally approach these problems?  Thanks.
 A: Here is an example of usage
$$
(x+1+O(x^{-1}))^x=
$$
$$
\begin{align}
&=x^x(1+x^{-1}+O(x^{-2}))^x\\
&=x^x\exp(x\log(1+x^{-1}+O(x^{-2})))\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-0.5(x^{-1}+O(x^{-2}))+o(x^{-1}+O(x^{-2}))^2))\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-0.5x^{-2}-x^{-1}O(x^{-2})-O(x^{-2})^2+o(x^{-1}+O(x^{-2}))^2))\tag{1}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-x^{-1}O(x^{-2})-O(x^{-2})^2+o(x^{-1}+O(O(x^{-1})))^2))\tag{3,4}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-x^{-1}O(x^{-2})-O(x^{-4})+o(x^{-1}+O(x^{-1}))^2))\tag{5,7}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-O(x^{-3})-O(x^{-4})+o(x^{-1})^2))\tag{9,12}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-O(O(x^{-2}))-O(O(x^{-2}))+o(x^{-1})^2))\tag{3}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})-O(x^{-2})-O(x^{-2})+o(x^{-2})))\tag{7}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})+o(x^{-2})))\tag{13}\\
&=x^x\exp(x(x^{-1}+O(x^{-2})))\tag{8}\\
&=x^x\exp(1+O(x^{-1}))\tag{12}\\
&=x^x\exp(1)\exp(O(x^{-1}))\\
&=x^x\exp(1)(1+O(x^{-1})+o(O(x^{-1})))\tag{2}\\
&=x^x\exp(1)(1+O(x^{-1}))\tag{10}\\
&=x^x e(1+O(x^{-1}))\\
&=ex^x+ex^xO(x^{-1})\\
&=ex^x+eO(x^xx^{-1})\tag{12}\\
&=ex^x+O(x^{x-1})\tag{11}\\
\end{align}
$$
In this solution I've used the following identites
$$
\begin{align}
\log(1+f)&=f-0.5f^2+o(f)^2\qquad&(1)\\
\exp(f)&=1+f+o(f)\qquad&(2)\\
x^{-m-n}&=O(x^{-m})\qquad&(3)\\
C\cdot f+O(f)&=O(f)\qquad&(4)\\
O(f)\cdot O(g)&=O(fg)\qquad&(5)\\
o(f)\cdot o(g)&=o(fg)\qquad&(6)\\
O(O(f))&=O(f)\qquad&(7)\\
O(f)+o(f)&=O(f)\qquad&(8)\\
o(f+O(f))&=o(f)\qquad&(9)\\
O(f)+o(O(f))&=O(f)\qquad&(10)\\
C\cdot O(f)&=O(f)\qquad&(11)\\
f\cdot O(g)&=O(f\cdot g)\qquad&(12)\\ 
C\cdot O(f)+D\cdot O(f)&=O(f)\qquad&(13)\\
\end{align}
$$
where $f=o(1)$
A: Using the power series approxiations
$$
\begin{align}
\log(1+t)&=t+O(t^2)\\
\exp(t)&=1+t+O(t^2)
\end{align}
$$
as $t\to0$, we get, as $x\to\infty$,
$$
\begin{align}
\left(1+\frac{1}{x}+O\left(x^{-2}\right)\right)^x
&=\exp\left(x\log\left(1+\frac{1}{x}+O\left(x^{-2}\right)\right)\right)\\
&=\exp\left(x\left(\frac{1}{x}+O\left(x^{-2}\right)\right)\right)\\
&=\exp\left(1+O\left(x^{-1}\right)\right)\\
&=e\,\exp\left(O\left(x^{-1}\right)\right)\\
&=e\,\left(1+O\left(x^{-1}\right)\right)\\
&=e+O\left(x^{-1}\right)\tag{1}
\end{align}
$$
Next, we have
$$
\begin{align}
\left(x+1+O\left(x^{-1}\right)\right)^x
&=ex^x+x^x\left(\left(1+\frac{1}{x}+O\left(x^{-2}\right)\right)^x-e\right)\\
&=ex^x+x^x\left(e+O\left(x^{-1}\right)-e\right)\\
&=ex^x+O\left(x^{x-1}\right)\tag{2}
\end{align}
$$
A: What about this slick one:
$$(x+1+O(x^{-1}))^x = (x(1 + x^{-1}) +O(x^{-1}))^x$$
$$=O(x^x(1+\frac{1}{x})^x) + O(x^{x-1}) = ex^x + O(x^{x-1})$$
TA-DA!!!
