Partial Fraction Decomposition?

For repeated, linear factors, (the term $(x-r)^m$ appears in the denominators for some integer $r$ and $m$, where $m > 1$) there is a partial fraction for each power upto and including $m$. Why?

Why can't you just have one partial fraction for the term with the $m^{th}$ power? Could someone give me an explanation about why you need a partial fraction for each power?

Hint $\ $ Every proper fraction of form $f(x)/x^m$ can be written as $(c_0 + c_1 x +\:\cdots\: c_{m-1} x^{m-1})/x^{m}$. Such proper fractions form an $m$ dimensional vector space over $\mathbb R$ with basis $x^{-1},\:x^{-2},\:\cdots\:x^{-m}\:$. Therefore if $\,m > 1\,$ then the single fraction $1/x^m$ does not span the whole space. The same holds true if one applies a shift automorphism $x\mapsto x-r$. More concretely, e.g.

$$\dfrac {1}{x-r} = \dfrac{a}{(x-r)^m} \iff (x-r)^m = a(x-r) \iff m = 1 = a$$

Or, examine below how Heaviside breaks down when $\,a= b$

$$\dfrac1{(\color{#c00}{x\!-\!a})(\color{#0a0}{x\!-\!b})}=\dfrac A{x-a}+\dfrac B{x-b}\,\ \Rightarrow\ \, A = \dfrac{1}{a\color{#0a0}{-b}},\,\ B =\dfrac{1}{b\color{#c00}{-a}} =\, {-}A$$

Think about what happens with an irreducible quadratic, say $x^2+1$ in the denominator. You have to setup the form as $\dfrac{Ax+B}{x^2+1}$. The degree in the numerator is always one less than the degree in the denominator. So you can technically setup what you want with a single fraction but it isn't $\dfrac{A}{(x-r)^m}$. It would have to be $\dfrac{A_{m-1}x^{m-1}+...+A_1x+A_0}{(x-r)^m}$. This is a big mess to deal with in solving the rest of the decomposition. By clever rewriting of the numerator you can get $B_{m-1}(x-r)^{m-1}+...+B_1(x-1)+B_0$. This then allows you to write the partial fraction as separate pieces with constants on top.