Here is another solution.
Claim 1: If $f:S^n \to S^n$ has degree $d$ then so does $\Sigma f: S^{n+1} \to S^{n+1}$
Proof: Use the Mayer-Vietrois sequence for $S^{n+1}$. Let $A$ be the complement of the North pole, and $B$ the complement of the South pole. Then $S^n \simeq A \cap B$ and the connecting map $\partial_*$ in the Mayer-Vietrois sequence is an isomorphism. We get the following commutative diagram
$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} H_{n+1}(S^{n+1}) & \ra{\partial_*} & H_n\left(A \cap B\right) & \la{i_*} & H_n(S^n)\\ \da{\Sigma f_*} & & \da{} & & \da{f_*} \\ H_{n+1}(S^{n+1}) & \ra{\partial_*} & H_n\left(A \cap B\right) & \la{i_*} & H_n(S^n)\\ \end{array} $
in which each horizontal map is an isomorphism. Thus $\Sigma f_* = \partial_*^{-1} i_* f_* i_*^{-1}\partial_*$ and applying homology shows that $\text{deg}(f) = \text{deg}(\Sigma f)$
Thus we are reduced to simply showing that there is a map $f:S^1 \to S^1$ of degree $k$. But this is just the winding number and it is (reasonably well known) that the map $z \mapsto z^k$ (where we view $S^1$ as the unit circle in $\mathbb{C}$) has degree $k$.
Finally I would direct you to have a look at Algebraic Topology by Hatcher:
- Example 2.31 gives a direct construction of a map of arbitrary degree;
- Example 2.32 works through the calculation of the map $f(z)=z^k$ proving it has degree $k$; and
- Prop 2.33 gives another prove of Claim 1 above (which basically takes a different route to the same commutative diagram).