Use space suspension and reduced homology to compute the homology group $H_{k}(\mathbb{S}^{n})$ for any $n\geq 0$. For a space $X$, the suspension of it is defined by $$\Sigma X:=C_{+}(X)\cup_{X} C_{-}(X),$$ where $C_{+}(X)$ and $C_{-}(X)$ are the upper and lower cone of it.
The reduced homology, roughly speaking, it is the homology obtained by removing $H_{k}(\text{point})$ from $H_{k}(X)$. 
For a  usual space, this means that 
$$\overline{H}_{k}(X)=\left\{
     \begin{array}{ll}
     H_{k}(X),\ \ \text{for}\ k>0\\
     \\
     \mathbb{Z}^{\#\ \text{of path components-1}},\ \ \text{for}\ k=0.
     \end{array}
    \right.
$$
A well-known theorem and corollary are as follows:

Theorem. $\overline{H}_{k+1}(\Sigma X)\cong \overline{H}_{k}(X)$.
Corollary: $\overline{H}_{k}\Big(\Sigma^{\ell}X\Big)\cong \overline{H}_{k-\ell}(X).$

Now, I want to apply these two results to compute the homology group of $\mathbb{S}^{n}$. I computed the case of $k=n$ as follows:
Recall from the first theorem that $\overline{H}_{k+1}(\Sigma X)\cong \overline{H}_{k}(X)$.
For $X=\mathbb{S}^{n}$, we know that $\Sigma X=\mathbb{S}^{n+1}$, so the above identity gives us $$\overline{H}_{k+1}(\mathbb{S}^{n+1})\cong\overline{H}_{k}(\mathbb{S}^{n}).$$
For $k=n$, we firstly consider the case of $n=0$ where $\mathbb{S}^{0}$ is two points,  then $$\overline{H}_{0}(\mathbb{S}^{0})=\mathbb{Z}^{\#\ \text{of path components of}\ \mathbb{S}^{0}}=\mathbb{Z},$$ and thus $$H_{1}(\mathbb{S}^{1})=\overline{H}_{1}(\mathbb{S}^{1})=\overline{H}_{0}(\mathbb{S}^{0})=\mathbb{Z}.$$
Then consider $n=1$, we have $$H_{2}(\mathbb{S}^{2})=\overline{H}_{2}(\mathbb{S}^{2})=\overline{H}_{1}(\mathbb{S}^{1})=\mathbb{Z}.$$ 
Keep iterating, an inductive argument will show that $H_{k}(\mathbb{S}^{n})=\mathbb{Z}$ for $k=n$. 

But then I got stuck, when I tried to compute in the case of $k=0$.
Firstly, from the case of $k=n$, we see that $H_{0}(\mathbb{S}^{0})=\mathbb{Z}$, but then for $n=1$, we have we have $$\overline{H}_{1}(\mathbb{S}^{2})=\overline{H}_{0}(\mathbb{S}^{1}),$$ what should I do next without knowing the number of path-components of $\mathbb{S}^{n}$ for $n\geq 1$? I understand that $\mathbb{S}^{n}$ is path-connected, but I don't know how to start from this to compute the number of path-components. 
If path-connectedness implies there is one path-components, then okay no problem we have $$H_{0}(\mathbb{S}^{n})=\mathbb{Z}.$$ But then do we have contradiction? since then we have $$H_{1}(\mathbb{S}^{n})=\overline{H}_{1}(\mathbb{S}^{n})=\overline{H}_{0}(\mathbb{S}^{n})=\mathbb{Z}$$ but we know that $H_{k}(\mathbb{S}^{n})=0$ for $0<k<n$.
so path-connected means there does not exist path-component? and thus $$H_{0}(\mathbb{S}^{n})=0$$ but then what should I do to compute $H_{0}(\mathbb{S}^{n})$ for $n>0$?
There seems no other way to iterate...

Similarly problem also arose when I tried to compute for $k<n$. If we know the case of $k=0$, then we can separate the case where $k=0<n$, and for all other $1\leq k<n$, $$\overline{H}_{k}(\mathbb{S}^{n})=H_{k}(\mathbb{S}^{n})$$ but what should I do to compute this homology group if I don't know $H_{k}(\mathbb{S}^{n})=0$ for $0<k<n$??

I am kind of believing that there is no way to compute the homology using only these two result other than the case of $k=n$...
Thank you for any advice and help!

Edit 1:
Okay, I figured it out. This can be done using induction. See my answer.
 A: First note that for all $n\geq 1$, $H_{0}(\mathbb{S}^{n})=\mathbb{Z}$ since $\mathbb{S}^{n}$ is path-connected for all $n\geq 1$.
To compute the higher dimensions, we use induction on $n$. Recall from the first theorem that $\overline{H}_{k+1}(\Sigma X)\cong \overline{H}_{k}(X)$.
Suppose we know the homology group for $S^{n-1}$ for some fixed $n> 1$, which is 
$$H_{k}(\mathbb{S}^{n-1})=\left\{
     \begin{array}{ll}
     \mathbb{Z},\ \ \ k=n-1\\
     0,\ \ \ 1\leq k\leq n-2\\
     \mathbb{Z},\ \ \  k=0.
     \end{array}
    \right.
$$
This theorem implies that 
$$\overline{H}_{k}(\mathbb{S}^{n-1})=\left\{
        \begin{array}{ll}
        H_{k}(\mathbb{S}^{n-1}),\ \ \ k>0\\
        \\
        \mathbb{Z}^{\{\#\ \text{of path components}-1\}},\ \ \ k=0
        \end{array}
       \right.
=\left\{
 \begin{array}{ll}
 \mathbb{Z},\ \ \ k=n-1\\
 0,\ \ \ 1\leq k\leq n-2\\
 0,\ \ \ k=0
 \end{array}
\right. 
$$
Therefore, $$\overline{H}_{k+1}(\mathbb{S}^{n})=\overline{H}_{k+1}(\Sigma \mathbb{S}^{n-1})=\overline{H}_{k}(\mathbb{S}^{n-1})=\left\{
 \begin{array}{ll}
 \mathbb{Z},\ \ \ k=n-1\\
 0,\ \ \ 1\leq k\leq n-2\\
 0,\ \ \ k=0
 \end{array}
\right. 
=\left\{
 \begin{array}{ll}
 \mathbb{Z},\ \ \ k+1=n\\
 0,\ \ \ 2\leq k+1\leq n-1\\
 0,\ \ \ k+1=1
 \end{array}
 \right.
$$
Therefore, $H_{k}(\mathbb{S}^{n})$ is known, since for all $k>0$, the reduced homology is the same as the regular topology, and we know what $k=0$ is, as mentioned as beginning.
$$H_{k}(\mathbb{S}^{n})=\left\{
      \begin{array}{ll}
      \mathbb{Z}, k=n\\
      0, 1\leq k\leq n-1\\
      \mathbb{Z}, k=0.
      \end{array}
     \right.
$$
Doing this inductively, this trick is actually doing this below graph:

Thus, the only thing you need to do is to compute $H_{k}(\mathbb{S}^{1})$ to use the inductive graph and $H_{k}(\mathbb{S}^{0})$ as a special case.
The computation of $H_{k}(\mathbb{S}^{1})$ is regular. To compute $H_{k}(\mathbb{S}^{0})$, note that $\mathbb{S}^{0}$ is just two points. I claim that 
$$H_{k}(2\ pts)=H_{k}(pt)\oplus H_{k}(pt)=\left\{
         \begin{array}{ll}
         \mathbb{Z}\oplus\mathbb{Z},\ \ \ k=0\\
         0,\ \ \ k>0
         \end{array}
        \right.
$$
This can be proved easily using excision isomorphism: $$H_{k}(2pts, pt)\cong_{excision}H_{k}(pt,\varnothing)=H_{k}(pt),$$ and then just do the long exact sequence thing. 
