Finding $E(X^Y)$ If $X,Y$ are jointly distributed discrete random variables with pmf,
$p_{X,Y,}(x,y)=\frac{\theta}{\exp{\theta}}\frac{(1-\theta)^x\theta^y}{y!}\:x,y=0,1,2,3...\:, 0\lt\theta\lt1$.
Find $E(X^Y)$.
Ok, so I end up with , $\frac{\theta}{(\exp{\theta})(1-(1-\theta)\exp{\theta})}$, after working through definitions, is this correct?

The expectation of $X^Y$ is 
$$\frac{\theta}{\exp(\theta)}\sum_x \sum_y x^y \frac{(1-\theta)^x \theta^y}{y!}.$$
Bring the $x^y$ part next to the $\theta^y$, where it wants to be.  Note that their product is $(x\theta)^y$. So our expectation is 
$$\frac{\theta}{\exp(\theta)}\sum_x (1-\theta)^x\left(\sum_y  \frac{(x\theta)^y}{y!}\right).$$
We recognize the inner sum as $e^{x\theta}$, so now we need
$$\frac{\theta}{\exp(\theta)}\sum_x (1-\theta)^xe^{x\theta}.$$
Note that $(1-\theta)^x e^{x\theta}=((1-\theta)e^\theta)^x$. Thus the above sum is the sum of a geometric series with common ratio $(1-\theta)e^\theta$. Thus our expectation is 
$$\frac{\theta}{\exp(\theta)}\cdot\frac{1}{1-((1-\theta)e^\theta)},$$
which is a version of what you wrote.
Added: As pointed out by Alex,  one should verify that $0\le (1-\theta)e^\theta \lt 1$, to ensure that the geometric series converges. One way to do this is to note that the derivative of $(-\theta)e^\theta$ is $-\theta e^\theta$, which is negative for $\theta\gt 0$. The function $(1-\theta)e^\theta$ is $1$ at $0$, and $0$ at $1$, and decreasing in the interval $[0,\infty)$, so in particular $0\lt (1-\theta)e^\theta\lt 1)$ in the interval $(0,1)$. 
