Is it true that $\frac{\mathbb{C}\{z\}}{(z-\alpha)^n}$ is a $\mathbb{C}$-vector space of dimension $n$? 
Let $\alpha\in\mathbb{C}$, and let $\mathbb{C}\{z\}$ denote the ring of convergent power series in $z$ around $\alpha$. Consider the ideal $(z-\alpha)^n$ in $\mathbb{C}\{z\}$. Is it true that $\frac{\mathbb{C}\{z\}}{(z-\alpha)^n}$ is a $\mathbb{C}$-vector space of dimension $n$, with basis $1, (z-\alpha),\dots , (z-\alpha)^{n-1}$? 

What goes into proving this fact?
Thanks in advance.
 A: Well, $\;V:=\Bbb C\{z\}/\langle (z-\alpha)^n\rangle\;$ is definitely a ring and, thus, an abelian group. It's not hard to check it is also a $\;\Bbb C\,-$ module, which already turns it into a complex vector space.
Now, if for some $\;c_1,...,c_{n-1}\in\Bbb C\;$ we have that
$$c_1(z-\alpha)+\langle (z-\alpha)^n\rangle+c_2(z-\alpha)^2+\langle (z-\alpha)^n\rangle+\ldots+c_{n-1}(z-\alpha)^{n-1}+\langle (z-\alpha)^n\rangle=\overline 0$$
then
$$\sum_{k=1}^{n-1}c_k(z-\alpha)^k\in\langle (z-\alpha)^n\rangle\iff\sum_{k=1}^{n-1}c_k(z-\alpha)^k=f(z)(z-\alpha)^n\;,\;\;f(z)\in\Bbb C\{z\}$$
which is absurd as the left side has a zero of order at most $\,n-1\,$ at $\;z=\alpha\;$ , whereas the right hand has a zero at the same point of order at least $\;n\;$ (why?) .
Thus, the vectors $\;(z-\alpha),...,(z-\alpha)^{n-1}\;$ are $\;\Bbb C\,-$ linearly independent in $\;V\;$.
Finally, any element in $\;V\;$ is an expression of the form $\;f(z)+\langle(z-\alpha)^n\rangle\;$ .But we can write
$$f(z)=(z-\alpha)^kg(z)\;,\;\;k\in\Bbb N\cup\{0\}\;,\;\;\text{with}\;\;g(z)\in V\;\;\text{that doesn't vanish at}\;\;\alpha$$  and thus...well, try to take it from to show that group of vectors is a generator of $\;V\;$...but you seem to be missing the vector $\;1\;$ .
