Assume the Exchange Theorem and prove the following: Assume the vector space V is finitely generated. Then there is a natural number n such that the length of a linearly independent sequence is less than or equal to n.
Theorem: Assume V is spanned by a sequence $(v_1, . . . , v_n)$ of length n. Then any sequence $(w_1, . . . , w_{n+1})$ of length $n + 1$ is linearly dependent.
Approach
Proof
Let x be the cardinality of a linearly independent sequence
Assume a vector space V is finitely generated by a spanning sequence $B=(v_1,...,v_k)$. By the exchange theorem, if we append a $v_j \in V\B$, $B$ becomes linearly dependent, so there doesn't exist a linearly independent sequence of cardinality greater than $k$.
If $B$ is linearly independent $|B|=k=x$.
If $B$ is linearly dependent then,there exists maximal linearly independent spanning sub-sequence $(v_1,..,v_j)$ for $j<k$,so $j=x$
How does that look?