ok, let's start from the basics and build our way up.
the factorial of an integer is defined as the product of all integers starting from 1 going up to the integer.
example, 5! = 1*2*3*4*5 and 6! = 1*2*3*4*5*6
so the integers go up in order until we get to the integer we are finding the factorial of. but how do we go from one integer to the other? we simply add 1 to the current integer to get the next integer. equivalently, this means we can subtract one from an integer to get the previous integer and keep doing so until we hit 1 and stop, which is what we do here. take 5! again as an example.
we can say that 5! = 5*4*3*2*1, the same thing, just going backwards. since we are going backwards, we subtract 1 from the first integer to get the second, and then subtract 1 from the second integer to get the third and so on. so we can say,
5! = 5*(5 - 1)*((5 - 1) - 1)*(((5 - 1) - 1) - 1)*((((5 - 1) - 1) - 1) - 1)
which is just simpler to write as
5! = 5*(5 - 1)(5 - 2)(5 - 3)(5 - 4)
if the factorial is very long or indefinite, we can omit some of the terms in the middle, again, let's use 5! as an illustration
5! = 5*(5 - 1)*...*(5 - 3)(5 - 4)
but we usually simplify the end, so instead of 5 - 4 we write 1 and instead of 5 - 3 we write 2, so we get:
5! = 5*(5 - 1)*...*2*1
but this representation can work for any integer. if we want to have an arbitrary integer factorial, we can use n. so in general:
n! = n(n - 1)(n - 2)(n - 3)....(3)(2)(1)
there is no difference if we want (n + 1)!, we just start at n + 1 and subtract 1 until we get to 1
(n + 1)! = (n + 1)(n)(n - 1)(n - 2)...(3)(2)(1)
so here is where the algebraic manipulation comes in.
recall that n! = n(n - 1)(n - 2)(n - 3)....(3)(2)(1)
now let's write (n + 1)! out.
we have (n + 1)! = (n + 1)(n)(n - 1)(n - 2)(n - 3)...(3)(2)(1)
notice anything? yes, what's in red is n! so we can just replace it
thus we get, (n + 1)! = (n + 1)n!
do you get it now?