Does anyone understand this calculus solution

The following is the textbook solution to my problem (I use {} to indicate subscript):
a{n+2} = -a{n}/(n+1)
n = 0 => a{2} = -a{0}
n = 2 => a{4} = -a{2}/3 = a{0}/(1*3)
n = 4 => a{6} = -a{4}/5 = -a{0}/(1*3*5)
n = 6 => a{8} = -a{6}/7 = a{0}/(1*3*5*7)
etc.
the emerging pattern indicates
a{2k} = (-1)^k * a{0} / (1*3*5*...(2k-1))
= (-1)^k * 2*4*6...(2k)a{0} / (2k)!
= (-1)^k * 2^k * k! * a{0} / (2k)!

I understand up until the last line. how does 2 * 4 * 6 ... (2k) equal 2^k k!
i can see how 2 k! would work but not 2^k

If I take the above line n = 6 => a{8} = -a{6}/7 = a{0}/(1*3*5*7) and use a{2k} = (-1)^k * 2^k * k! * a{0} / (2k)! I don't get the same thing
a{8} = a{2k} = (-1)^k * 2^k * k! * a{0} / (2k)!
= a{2*4)} = (-1)^4 * 2^4 * 4! * a{0} / (2*4)!
a{8} = a{8} = 2*2*2*2 * 4*3*2*1 * a{0} / (8*7*6*5*4*3*2*1)
a{8} = a{8} = 2*2*2 * 8*6*4*2 * a{0} / (8*7*6*5*4*3*2*1)
a{8} = a{8} = 2*2*2 * a{0} / (7*5*3*1)
which does not equal a{0}/(1*3*5*7)
if it were 2 as opposed to 2^k this would work out. I don't think this is a misprint because it is not the first time am seeing this occur.

2 * 4 * 6 * ... * (2k)
= (1*2) * (2*2) * (3*2) ... * (k*2)
= (2 * 2 * 2 * ... * 2) * (1 * 2 * 3 * ...k)
= 2^k * k!

.

You made a mistake going from this...
a{8} = a{8} = 2*2*2*2 * 4*3*2*1 * a{0} / (8*7*6*5*4*3*2*1)
to this...
a{8} = a{8} = 2*2*2 * 8*6*4*2 * a{0} / (8*7*6*5*4*3*2*1)
You have multiplied four of the factors by 2 so you should have removed all of the twos, not just one of them.