From Rotman's Algebraic Topology
If $f \colon S^{2n} \rightarrow S^{2n}$, then either $f$ has a fixed point or some point is sent to its antipode.
The second half of the proof is as follows:
Suppose that $f(x) \neq -x, \forall x \in S^{2n}$. Define $g(x) = -f(x)$. THen $g$ has no fixed points, and so $-f = g \simeq a^{2n}$, where $a^n$ represents the antipodal map. Then $f \simeq -a^{2n} = 1_{S^{2n}}$ and $d(f) = 1$, a contradiction.
What does $d(f) = 1$ contradict?
In his proof, Rotman shows first that $d(f)=-1$ by Theorem 6.23; this is what $d(f)=1$ contradicts.