$K^n \cong K^m \implies n = m$ Let $K$ be a field and let $E$ be a vector space over $K$. I want to prove that any two finite bases of $E$ are equinumerous. What I did was:
Let $B = \{u_1, \cdots, u_n\}$ be a finite basis of $E$. Consider:
$$f: K^n \longrightarrow E$$
Defined by $f(\alpha_1, \cdots, \alpha_n) = \sum_{i=1}^n\alpha_i u_i$
It can be easily shown that $f$ is an isomorphism. It follows that $K^n \cong E$. 
Let $B_0$ be any other finite basis of $E$, say of cardinality $m$. In a similar reasoning to above, we can prove that $K^m \cong E$.
It follows that $K^n \cong K^m$. 
Now, how do I show that $n = m$?
I know that an isomorphism preserves the algebraic structure, that's, if two objects are isomorphic, then what is true about a designated part of one object's structure is true about the other's. That's why one feels that $n = m$ should be true. But how do I prove that rigorously?
Thanks a lot.
 A: Although this question has essentially already been answered in the comments, I will present a solution anyway.  It's obvious that $K^n$ has a basis of cardinality $n$, namely $\{(1,0,\ldots,0), (0,1,0,\ldots,0), \ldots, (0,\ldots,0,1)\}$, so to prove that $K^n \cong K^m$ implies $n = m$, it's enough to prove that any two finite bases of a vector space have the same cardinality.  The simplest argument to show this that I know of involves successively exchanging elements of one basis with the other and is given in the Wikipedia articles http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces#Alternative_Proof and http://en.wikipedia.org/wiki/Steinitz_exchange_lemma.  
To illustrate this proof by example, suppose that the vector space $V$ had a basis $B=\{v,w,x\}$ of cardinality $3$ and a basis $B'=\{a,b,c,d,e\}$ of the larger cardinality $5$.  Then, since $B$ is a basis, $a$ must be a linear combination  of $v$, $w$, and $x$: 
$$\alpha v + \beta w + \gamma x-a=0.\qquad (1)$$
 Also, since $B'$ is a basis, $\{a\}$ must be linearly independent, which means that $a\ne 0$.  Therefore, at least one of $\alpha$, $\beta$, and $\gamma$ must be nonzero; this means that we can rewrite $(1)$ to express one of the elements of $B$ as a linear combination of the other elements of $B$ together with $a$.  Assuming without loss of generality that the nonzero coefficient is $\alpha$, we can express $v$ 
 as a linear combination of $w$, $x$ and $a$:
$$
v=-\alpha^{-1}\beta w-\alpha^{-1}\gamma x+\alpha^{-1} a.
$$  Therefore, since $B=\{v,w,x\}$ spans $V$, $\{w,x,a\}$ must also span $V$.  
We can now perform the same step again: write $b$ as a linear combination of $w$, $x$ and $a$:
$$\epsilon w + \zeta x + \mu a-b=0.$$
Since $\{a,b\}$ is a subset of the basis $B'$, it is linearly independent, so either $\epsilon$ or $\zeta$ must be nonzero.  This means that we can express either $w$ or $x$ as a linear combination of the other elements of $\{w,x,a\}$ together with $b$.  Assuming again without loss of generality that we can express $w$ in this way, $w$ is a linear combination of $x$, $a$ and $b$, and so $\{x,a,b\}$ must span $V$.  
Performing the same step one more time, in the same way, we find that $\{a,b,c\}$ must span $V$.  Therefore, $d$ is a linear combination of $a$, $b$, and $c$, which contradicts the linear independence of $B'$ and hence the assumption that $B'$ was a basis.  Since we can clearly perform a similar argument for any two finite bases of unequal cardinalities, we can conclude that any two finite bases of a vector space must have the same size.
