Atiyah-Macdonald, Exercise 8.3: Artinian iff finite k-algebra. 
Atiyah Macdonald, Exercise 8.3. Let $k$ be a field and $A$ a finitely generated $k$-algebra. Prove that the following are equivalent:
  (1) $A$ is Artinian.
  (2) $A$ is a finite $k$-algebra.

I have a question in the proof of (1$\Rightarrow$2): By using the structure theorem, we may assume that $(A,m)$ is an Artin local ring. Then $A/m$ is a finite algebraic extension of $k$ by Zariski lemma. Since $A$ is Artinian, $m$ is the nilradical of $A$ and thus $m^n=0$ for some $n$. Thus we have a chain $A \supseteq m \supseteq m^2 \supseteq \cdots \supseteq m^n=0$. Since $A$ is Noetherian, $m$ is finitely generated and hence each $m^i/m^{i+1}$ is a finite dimensional $A/m$-vector space, hence a finite dimensional $k$-vector space.
But now how can I deduce that $A$ is a finite dimensional $k$-vector space?

The claim also seems to follow from the Noether normalization lemma:
Let $B := k[x_1, \dotsc, x_n]$ with $k$ any field and let $I \subseteq B$ be any ideal.
Since $A$ is a finitely generated $k$-algebra you may let $A := B/I$. By the Noether normalization lemma it follows that there is a finite set of elements $y_1, \dotsc, y_d \in A$ with $d = \dim(X)$ and the property that the subring $k[y_1, \dotsc, y_d] \subseteq A$ generated by the elements $y_i$ is a polynomial ring. The ring extension $k[y_1, \dotsc, y_d] \subseteq A$ is an integral extension of rings. If $d  = 0$, it follows from the same lemma the ring extension $k \subseteq A$ is integral and since $A$ is finitely generated as $k$-algebra by the elements $\overline{x_i}$ and since each element $\overline{x_i}$ is integral over $k$, it follows that $\dim_k(A) < \infty$.
Question: “But now how can I deduce that $A$ is a finite dimensional $k$-vector space?”
Answer: It seems from the argument above you can use the Noether normalization lemma to give another proof of your implication, different from the proofs given above. Hence now you have two proofs of your result.
