Proof:

We take $U = S^n - \{{\bf e}_{n+1}\}$ and
$V = S^n - \{-{\bf e}_{n+1}\}$
and note that $U\cap V$ deformation retracts to $S^{n-1}$. Consider the portion of the Mayer Vietoris sequence

$$\begin{CD} @> >> H_n(U)\oplus H_n(V) @> >> H_n(U\cup V) @>{\delta_n} >> H_{n-1}(U\cap V) @> >> H_{n-1}(U)\oplus H_{n-1}(V) @> >> \end{CD}$$

Since $U$ and $V$ are contractible spaces, we get for the case $n\geq 2$,

$$\begin{CD} 0 @> >> H_n(U\cup V) @>{\delta_n} >> H_{n-1}(U\cap V) @> >> 0, \end{CD}$$

and hence $H_n(S^n)\cong H_{n-1}(S^{n-1})$ ($n\geq 2$).
By induction the result would follow as soon as we prove it for the case $n = 1$.
For this case let us take a look at the end of the Mayer Vietoris sequence:

$$\begin{CD} 0 @> >> H_1(S^1) @> \delta_1 >> H_0(U\cap V) @>(\kappa... ...\kappa^{\prime\prime}) >> H_0(U)\oplus H_0(V) @> >> H_0(U\cup V) @> >> \end{CD}$$

Since $\delta_1$ is injective,

$$H_1(S^1) \cong$$   im$$\;\delta_1 =$$   ker$$\;(\kappa^{\prime}, -\kappa^{\prime\prime}).$$

To understand the map $(\kappa^{\prime}, -\kappa^{\prime\prime})$ we take a basis of $H_0(U\cap V)$ consisting of a pair of points $a \in U$ and $b \in V$.
The singleton $\{a\}$ generates $H_0(U)$ and

$$k^{\prime}(ma+nb) = ma + nb = n(b-a) + (m+n)a$$

which is a boundary in $H_0(U)$ if and only if $m+n = 0$.
Likewise $k^{\prime\prime}(ma+nb) = 0$ in $H_0(V)$ if and only if $m+n = 0$. Thus the kernel of $(\kappa^{\prime}, \kappa^{\prime\prime})$ is the infinite cyclic group
generated by the zero chain $a-b$. Hence we get

$$H_1(S^1) \cong \mathbb{Z}.$$

To calculate $H_n(S^1)$
for $n\geq 2$
we look at the portion of the Mayer Vietoris sequence

$$\begin{CD} @> >> H_n(U)\oplus H_n(V) @> >> H_n(U\cup V) @>{\delta_n} >> H_{n-1}(U\cap V) @> >> \end{CD}$$

and observe that since $H_n(U) = H_n(V) = H_{n-1}(U\cap V) = 0$ when $n\geq 2$,

$$H_n(S^1) = \{0\},\quad n\geq 2.$$