the question (from burton's elementary number theory);
$verify\ that\ \forall n\ge 1,$ $$2\cdot6\cdot10\cdots(4n-2)=\frac{(2n)!}{n!}$$
my work/proof;
this is obviously true for $n=1$, so assume this is true for $n\le k$
$$2\cdots(4k-2)=\frac{(2k)!}{k!}$$
now show the formula holds for $k+1$
$$2\cdots(4k-2)(4k+2)=\frac{(2k)!(4k+2)}{k!}$$
and this is where i feel like im cheating; because i know this will be true for $k+1$ when
$$\frac{(2k!)(4k+2)}{k!}=\frac{(2k+2)!}{(k+1)!}$$
but i didn't know how to go about it directly, so this is what i did;
$$\frac{(2k!)(4k+2)}{k!}=(k+1)(k+2)\cdots(2k)(4k+2)$$
and
$$\frac{(2k+2)!}{(k+1)!}=(k+2)(k+3)\cdots(2k)(2k+1)(2k+2)$$
now assuming that the formula holds for $k+1$, i assumed that the two expressions were equal, and setting them equal to each other everything but the following cancels;
$$(k+1)(4k+2)=4k^2+6k+2$$ and, $$(2k+1)(2k+2)=4k^2+6k+2$$ and clearly both are equal to each other and so it holds for $k+1$.
now i feel like this is cheating because, i had the formula to work with. i mean, i was able to demonstrate, i think, that the formula holds for $k+1$ but if i were asked to come up with $$2\cdot6\cdot10\cdots(4n-2)=\frac{(2n)!}{n!}$$ or even, $$\frac{(2k!)(4k+2)}{k!}=\frac{(2k+2)!}{(k+1)!}$$ this is all much less obvious and i would not have come up with or even seen this. thanks for your input.
Hint. For a simpler way, use the fact that $$4k+2=\frac{(2k+1)(2k+2)}{k+1}\ .$$
Your proof looks as if it should be ok, except that it is backwards - at one point you have assumed what you wish to prove - and you need to check that everything works "the right way around".
For an easier way to do the whole thing without explicitly using induction, $$\eqalign{2\times6\times10\times\cdots\times(4n-2) &=2^n\times1\times3\times5\times\cdots\times(2n-1)\cr &=2^n\times\frac{(2n)!}{2\times4\times6\times\cdots\times(2n)}\cr &=\frac{(2n)!}{n!}\ .\cr}$$