Why are there no poles for finite length signals
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.