 Solution

Expand $e_{r+1}$ in the given spanning set:Some $b_j$ must be nonzero, since otherwise $e_{r+1}$ would lie in the span of the earlier $e_i$, contradicting linear independence. Reorder the remaining $f_j$ so $b_{r+1}\ne 0$. Solving this equation for $f_{r+1}$ expresses it in terms of $e_1,\dots ,e_{r+1},f_{r+2},\dots ,f_m$. Every member of the old spanning set is therefore in the new span, so the replacement set still spans $V$. This proves the needed step of the Steinitz exchange lemma directly.

Starting with $f_1,\dots ,f_m$, apply that replacement successively to $e_1,e_2,\dots$. After $r\le m$ steps the spanning list is $e_1,\dots ,e_r$ together with $m-r$ unreplaced $f$'s. If $n>m$, after $m$ replacements the list $e_1,\dots ,e_m$ spans $V$, forcing $e_{m+1}$ to be a linear combination of it. That contradicts linear independence. Hence $\boxed{n\le m}$.