Proof of a Field Extensions Theorem Consider the following result.
Theorem : Let $E/F$ be a finite field extension of degree $n$ and let $V$ be a vector space over $E$. Then
$$
\dim_F V = [E:F] \dim_E V.
$$
Now, it seems like a standard proof of this theorem begins with :

Let $\mathcal{B}$ be a basis of $V$ over $E$ and $\mathcal{C}$ be a basis of $E/F$. We show that
  $$ W := \{ cb : c \in \mathcal{C}, b \in \mathcal{B} \} $$
  is a basis of $V$ over $F$.

Question : Would the following (shorter) proof be correct ?

Let $k := \dim_E V$. If $k$ is infinite then clearly $\dim_F V$ is infinite. Hence suppose $k < \infty$. Then
  $$V \cong \overbrace{E \oplus E \cdots \oplus}^{k \text{ times}} E =: E^{\oplus_k}$$
  and since
  $$E \cong F^{\oplus_{[E:F]}},$$
  we have
  $$V \cong \left( F^{\oplus_{[E:F]}} \right)^{\oplus_k} = F^{\oplus_{k [E:F]}}.$$


The proof is correct. But to be precise, you should always specify the category you are working in. For example, $E \cong F^n$ (where $n:=[E:F]$) is not correct as rings, but rather as $F$-vector spaces. Since $V \cong E^k$ as $E$-vector spaces also implies $V \cong E^k$ as $F$-vector spaces, it follows in fact $V \cong F^{nk}$ as $F$-vector spaces.
Notice that, going through the proof, this also produces an $F$-basis of $V$ out of an $F$-basis of $E$ and an $E$-basis of $V$ (by taking products). The common proof just writes down this set and proves that it is a basis, which is basically just a repetition of $(F^n)^k \cong F^{nk}$. It is good that you just use this general fact.
