= Solution
A <basis> of a <vector space> $V$ is a family of vectors that is both <linear independence>[linearly independent] and a <spanning set> of $V$.
Write the given finite basis as $\mathcal B=\{v_1,\ldots,v_n\}$. First, any other basis $\mathcal B'$ must also be finite. Each $v_i$ is a finite <linear combination> of elements of $\mathcal B'$, so the union $S$ of the finitely many elements of $\mathcal B'$ occurring in these $n$ expressions is finite. Since $S$ spans every $v_i$, it spans $V$. If some $w\in\mathcal B'\setminus S$ existed, then $w$ would lie in the span of $S$, contradicting the <linear independence> of $\mathcal B'$. Hence $\mathcal B'=S$; write $|\mathcal B'|=m$.
It remains to prove the elementary <Steinitz exchange lemma> directly. If independent vectors $w_1,\ldots,w_r$ lie in the span of $v_1,\ldots,v_n$, express $w_1$ in terms of the $v_i$. Some coefficient is nonzero, so the corresponding $v_i$ can be solved for in terms of $w_1$ and the other $v_i$; replacing it by $w_1$ preserves the span. Inductively, after replacing $k$ of the $v_i$ by $w_1,\ldots,w_k$, the expression for $w_{k+1}$ must have a nonzero coefficient on one of the unreplaced $v_i$, since otherwise $w_{k+1}$ would be a linear combination of $w_1,\ldots,w_k$. That $v_i$ can again be replaced. There are only $n$ original vectors to replace, so $r\leq n$.
Apply this argument first to the independent family $\mathcal B'$ and the spanning family $\mathcal B$ to obtain $m\leq n$, and then with the two bases interchanged to obtain $n\leq m$. Therefore
$$
\boxed{|\mathcal B'|=|\mathcal B|}.
$$
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