Degree of a non-surjective map $S^n\to S^n$ is zero

Choose \(p\in S^n\setminus f(S^n)\). Then \(f\) factors as \[ S^n \mapsvia{f_1} S^n\smts{p}\mapsvia{j} S^n, \] where \(j\) is the inclusion. Since \(S^n\smts{p}\cong\RR^n\) is contractible, \[ H_n(S^n\smts{p})=0. \] Hence the induced map on top homology factors through the zero group: \[ H_n(S^n)\mapsvia{(f_1)_*}0\mapsvia{j_*}H_n(S^n). \] Thus \(f_*=0\) on \(H_n(S^n)\cong\ZZ\). By the definition of degree, \(\deg f=0\).