<s>[INST] Mathematical Induction (product of $n$ consecutive numbers) Assumption:
$$(n+1)(n+2) \cdots (2n) = (2^n)\cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)$$
Prove for $n+1$:
$$(n+2)(n+3) \cdots (2(n+1)) = (2^{n+1}) \cdot 1 \cdot 3 \cdot 5 \cdots (2(n+1)-1)$$
Using the assumption, I divide both sides by $(n+1)$ and substitute RHS into my $n+1$ equation, however it does not equate.
 [/INST] Assumption
$$C(n) := (n+1)(n+2) \cdots (2n) = (2^n)\cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)$$
Proof
Basis:
$$2=2^1$$
Inductive step:
$$C(n+1):= (n+2)(n+3) \cdots (2(n+1)) = (2^{n+1}) \cdot 1 \cdot 3 \cdot 5 \cdots (2(n+1)-1)$$
$$We\ have\ to\ prove: C(n) \Rightarrow C(n+1)$$
$$\vdots$$
$$Little\ substitution:$$
$$(n+2)(n+3) \cdots (2n)(2n+1)(2n+2) = (n+1)(n+2) \cdots (2n) \cdot 2 \cdot (2n+1)$$
$$That\ leaves\ us\ with:$$
$$(2n+2) = 2 \cdot (n+1)$$
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