Here's my proof:
Base Case:
Show that P(1) is true:
n = 1
(4(1) - 2) = (2(1))! / (1)!
4 - 2 = 2! / 1
2 = 2
The base case holds.
Induction Step:
Show that for all natural numbers k, if P(k) is true, then P(k+1) is true.
Suppose that k is any arbitrary natural number such that
2 · 6 · 10 · 14 · ... · (4k - 2) = (2k)! / k! (Induction Hypothesis)
Want to show: P(k+1) is true.
2 · 6 · 10 · 14 · ... · (4k - 2) · (4(k+1) - 2) = (2(k+1))! / (k + 1)!
((2k)! / k!)(4(k+1) - 2) = (2(k+1))! / (k + 1)!
((2k)! / k!)(4(k+1) - 2) = ((2k + 2)(2k + 1)(2k)!) / ((k+1)k!)
((k!) / (2k)!)((2k)! / k!)(4(k+1) - 2) = (((2k+2)(2k+1)(2k)!) / ((k+1)k!))((k!) / (2k!))
(4(k+1) - 2) = (2(k+1)(2k+1)) / (k+1)
4k + 4 - 2 = 2(2k+1)
4k + 2 = 4k + 2
Therefore, P(k+1) is true.
Q.E.D.
I don't understand how this proof is equivalent to assuming 2 = -2,
then squaring both sides (2)^2 = (-2)^2,
so 4 = 4,
therefore 2 = -2
Q.E.D.
I think you have probably confused yourself by your proof approach. You have basically assumed the result and then reduced the equality to one you know to be true. This is not a correct proof. Instead, you should start with one side and show that it is equal to the other side. So: \begin{align*} 2\cdot 4\cdot 6\cdots\cdot (4k-2)\cdot (4(k+1)-2) &= \frac{(2k)!}{k!}\cdot (4(k+1)-2) \\ &= \frac{(2k)!}{k!}\cdot(4k+2) \\ &= \frac{2(2k+1)(2k)!}{k!} \\ &= \frac{2(k+1)(2k+1)(2k)!}{(k+1)k!} \\ &= \frac{(2k+2)(2k+1)(2k)!}{(k+1)!} \\ &= \frac{(2k+2)!}{(k+1)!}. \end{align*}