This is a question for class. My prof has given me some guidance but I can't wrap my mind around it.
My prof said I should start by considering the equation for skewness very generically, and use the moment generating function. He also gave me a demonstration of swapping the interior of the skewness equation:
Skew($X$) = $\left(\frac{X-\mu}{\sigma}\right)^3 = \left(\frac{X - E(X)}{\sqrt{Var(X)}}\right)^3$
Then I know some properties of an exponential distribution: $E(X) = \frac{1}{\lambda}, Var(X)=\frac{1}{\lambda^2}, SD(X)=\frac{1}{\lambda}$
So with that I can say that: $\frac{X-E(X)}{\sqrt{Var(X)}} = \frac{x-\frac{1}{\lambda}}{\frac{1}{\lambda}} = \lambda x -1$
therefore, $Skew(X) = E(\lambda x -1)^3$
let $\lambda X = Y$, and expand $(Y-1)^3$ to $Y^3-3Y^2+3Y-1$
Then, because I am given that the nth moment of an Expo(1) random variable is n!, I can say: $Y^3 = 3!, Y^2 = 2!, Y = 1!$
And therefore:
$Y^3-3Y^2+3Y-1 = 3! -3(2!) + 3(1!) - 1 = 6 - 6 + 3 - 1 =2$