Why are there no poles for finite length signals My professor recently said something along the lines of, "finite length signals can't have poles because poles correspond to recursive difference equations which correspond to infinite length signals." 

I can't really wrap my head around this. Can someone explain in a different or more simple way why finite length signals can't have poles? Would much prefer a simple, hand-wavy explanation that I can remember easily rather than a rigorous explanation/proof.
Hand-wavy? A rational fraction $$\frac{N(X)}{D(X)}$$
where $N(X)$ and $D(X)$ are polynomials can be written as some polynomial $P(X)$ plus a finite amount of terms of the shape:

$$\frac{a_i}{X-X_i}\,.$$

Let us say that a $0$ pole ($X_i=0$) is harmless, because it is 
 just some $X^{-1}$. So  on the one hand, you have the monomials $X^d$, $d\ge0$, from $P$ and possibly a $X^{-1}$. If you replace the $X$ by a $z^{-n}$, you will still have a finite number of $z$ to the power of some integer, which are just delays. 

On the other hand, if you have one non-zero pole:

$$\frac{1}{X-X_j} = \frac{1}{X}\left(1+ \left(\frac{X_j}{X}\right)^1+\left(\frac{X_j}{X}\right)^2+\cdots\right)$$
and replace the $X$ by some $z^{-n}$, you will get an infinite quantity of $z$ to the power of some integer, and thus an infinite impulse response (IIR). The complicated part is to show that several non-zero poles do not "cancel" each other. A lot of people confuse IIR and recursive.

Be cautious though that an apparently "recursive implementation" does not always imply non-zero poles: a FIR system with $z$-transform $1-z^{-2}+z^{-4}-z^{-6}$ can be written as: 

$$ \frac{1-z^{-8}}{1+z^{-2}}$$

that is, you can rewrite: $$y ( k ) = x ( k ) - x ( k - 2 ) + x ( k - 4 ) - x ( k - 6 )$$ as $$y ( k ) = x ( k ) - x ( k - 8 ) - y ( k - 2 )\,,$$

which is a recursive difference equation, yet with a FIR.