Induction is done by demonstrating that if the condition is true for some $n$ then it must also be true for $n+1$. If you then show that the condition is true for $n=0$ then it must be true for all $n>0$. For this problem:
Step $1$: $n=1$
The sum of the first $1$ odd numbers is $1$. $1^2=1$. Therefore the condition holds for $n=1$.
Step $2$: induction
If the sum of the first $n$ odd numbers is $n^2$ then the sum of the first $n+1$ integers is
$n^2 + (2n + 1) = (n+1)(n+1)=(n+1)^2$
So the condition is also true for $n+1$.
Step $3$: conclusion
Since the we have shown that the condition is true for $n=1$ and we have shown that if it is true for $n$ then it is also true for $n+1$ then it follows by induction that it is true for all $n\geq 1$.
General term: $s_n = \displaystyle \sum_{0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n-1-r}{r} \text{ for } n \in \mathbb{N} \tag{1}$
I think you struggled because the binomial sum $s_{n+2}$ has a decomposition that involves both its even and odd numbered predecessors, and not just any one of them, as you were trying in the inductive step.
Decomposition: $s_{n+2} = s_{n+1} + s_{n} \tag{2}$
This is easy to prove by using the combinatorial identity $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$. See here and here.
We can then use this property along with strong induction to prove:
$P(n): s_n = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n \text{ true } \forall n \in \mathbb{N}\tag{3}$
Proof-sketch:
Base case: Show $P(1)$ and $P(2)$ to be true (by evaluating both sides).
Inductive step: Assume $P(n)$ and $P(n+1)$ true.
Then $\begin{aligned}s_{n+2} & = s_{n+1} + s_{n} \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^{n+1} - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^{n+1} + \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^n\left(\frac{3 + \sqrt{5}}{2}\right) - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^n\left(\frac{3 - \sqrt{5}}{2}\right) \\ & = \frac{1}{\sqrt{5}}\left(\frac{1 + \sqrt{5}}{2}\right)^{n+2} - \frac{1}{\sqrt{5}}\left(\frac{1 - \sqrt{5}}{2}\right)^{n+2} \\ & \implies P(n+2) \text{ true }\end{aligned}$
This completes our proof. $ \blacksquare$
Note: The formulas for odd $n$ and even $n$ follow automatically from $(3)$ by the removal of the floor function in $s_n$.
Best Answer
The idea is to "complete the factorials":
$$ 1\cdot 3 \cdot 5 \cdots (2n-1) = \frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n-1)\cdot (2n) }{2\cdot 4 \cdot 6 \cdots (2n)} $$
Now take out the factor of $2$ from each term in the denominator:
$$ = \frac{ (2n)! }{2^n \left( 1\cdot 2 \cdot 3 \cdots n \right)} = \frac{(2n)!}{2^n n!}$$
A mathematician may object that there is a small gray area about what exactly happens between those ellipses, so for a completely rigorous proof one would take my post and incorporate it into a proof by induction.