Does 0.99999999999999999999999999999999999999999 repeating equals 1?

Yes.

The value 0.999999999999999999...
is the limit, as n goes to infinity, of a sequence of values S(n), where:
S(n) = Sum from k=1..n of 9 * (0.1)^k
This sequence S(n) converges to the limit 1.

Proof:

We know that
S(n) = Sum from k=1..n of 9 * (0.1)^k
= 9 * Sum from k=1..n of (0.1)^k
= 9 * { [ Sum from k=0..n of (0.1)^k ] - 1 }
It can be shown that
Sum from k=0..n of (0.1)^k
= [ 1 - (0.1)^(n+1) ] / [ 1 - 0.1 ]
= (10/9) * [ 1 - (0.1)^(n+1) ]

So
S(n) = 9 * { (10/9) * [ 1 - (0.1)^(n+1) ] - 1 }
= 10 * [ 1 - (0.1)^(n+1) ] - 9

As n increases the term (0.1)^(n+1) goes to zero.
Hence, S(n) converges to
= 10 * [ 1 - 0 ] - 9
= 1

To prove that

Sum from k=0..n of (0.1)^k
= [ 1 - (0.1)^(n+1) ] / [ 1 - 0.1 ]

First, let a = 0.1. Use induction.

If n = 0, then
Sum from k=0..n of a^k
= Sum from k=0..0 of a^k
= a^0
= 1
= [ 1 - a ] / [ 1 - a ]
= [ 1 - a^(0+1) ] / [ 1 - a ]
= [ 1 - a^(n+1) ] / [ 1 - a ]

Now, assume that
Sum from k=0..n of a^k
= [ 1 - a^(n+1) ] / [ 1 - a ]

and show that
Sum from k=0..(n+1) of a^k
= [ 1 - a^(n+2) ] / [ 1 - a ]

Well, we have:
Sum from k=0..(n+1) of a^k
= { Sum from k=0..n of a^k } + a^(n+1)
= [ 1 - a^(n+1) ] / [ 1 - a ] + a^(n+1)
= { 1 - a^(n+1) + [ 1 - a ] * a^(n+1) } / [ 1 - a ]
= [ 1 - a * a^(n+1) ] / [ 1 - a ]
= [ 1 - a^(n+2) ] / [ 1 - a ]

QED