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.
 A: My approach would be to "massage" the assumed equation to look more like the desired equation.
Hint:  Try multiplying the assumed equation by 2.  What's missing from the left side after this step?  What's missing from the right side?
A: 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:
$$C(n+1):= (n+2)(n+3) \cdots (2(n+1)) = (2^{n+1}) \cdot 1 \cdot 3 \cdot 5 \cdots (2(n+1)-1)$$
$$We\ have\ to\ prove: C(n) \Rightarrow C(n+1)$$
$$\vdots$$
$$Little\ substitution:$$
$$(n+2)(n+3) \cdots (2n)(2n+1)(2n+2) = (n+1)(n+2) \cdots (2n) \cdot 2 \cdot (2n+1)$$
$$That\ leaves\ us\ with:$$
$$(2n+2) = 2 \cdot (n+1)$$
A: HINT $\ $ Dividing the second equation by the  first yields the identity
$$\rm\frac{(2\:n+1)\ (2\:n+2)}{n+1}\ =\ \ 2\ (2\:n+1) $$
Thus the second equation is simply $\rm\ 2\ (2\:n+1)\ $ times the first equation.
Alternatively one can easily reduce the induction to a trivial induction that a product of 1's equals 1, see my prior posts on (multiplicative) telescopy.
