How does does exponent property work on $\left(xe^{\frac{1}{x}}-x\right)$ How does 
$\left(xe^{\frac{1}{x}}-x\right)$
become
$\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$
 A: $$\ xe^{\frac{1}{x}}-x=x(e^{\frac{1}{x}}-1)=\frac{e^{\frac{1}{x}}-1}{\frac{1}{x}}$$
Because when you have the product of two terms, say $\ a\cdot b$, it is equivalent to $\ \frac{1}{\frac{1}{a}}\cdot b=\frac{b}{\frac{1}{a}}$
A: Mosh answers your question from the correct order (perfect way), however I wanted to point out that you can go the other way too (and prove this). 
Here is how :
$$\frac{(e^{\frac{1}{x}}-1)}{(\frac{1}{x})}=(e^{\frac{1}{x}}-1)\div \frac{1}{x}=(e^{\frac{1}{x}}-1)× x= xe^{\frac{1}{x}}-x$$

Therefore:
$$\frac{(e^{\frac{1}{x}}-1)}{(\frac{1}{x})}=xe^{\frac{1}{x}}-x$$
A: With more intermediate steps :
$$\left(xe^{\frac{1}{x}}-x\right)=$$
$$=x\left(e^{\frac{1}{x}}-1\right)$$
Let $x=\frac{1}{t}$ Replace $x$ by $\frac{1}{t}$
$$=\frac{1}{t}\left(e^{\frac{1}{x}}-1\right)$$
$$=\frac{\left(e^{\frac{1}{x}}-1\right)}{t}$$
$t=\frac{1}{x}$ Replace $t$ by $\frac{1}{x}$
$$=\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$

Another nice method :
$$\left(xe^{\frac{1}{x}}-x\right)=$$
$$=\frac{x\left(e^{\frac{1}{x}}-1\right)}{1}$$
Mulptiply the numerator and the denominator by the same term $\frac{1}{x}$
$$=\frac{\frac{1}{x}x\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$
Simplify $\frac{1}{x}x=1$
$$=\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$
