0.9999.... (Q further in the text) is equal to 1 - we can prove it by proving that negation leads to contradiction.
Let's assume that Q != 1. That means a real number P must exist that falls between Q and 1, i.e. Q < P < 1 [1].
We know that P > Q. We will try to construct it's decimal representation, because any real number has a decimal representation (though transcendent real numbers have infinite non-periodic ones). The first digit must be 0 followed by decimal point - 1 and greater would mean P > 1 and -1 and smaller would mean P < Q, which would contradict the original assumption [1]. Lets find the second number. Clearly, it must be 9 - any other value would mean that P<Q (example - 0.7 < 0.999... ). This will be our first induction step [I1].
Then we will show that for n-th (n is a natural number) digit in decimal representation of Q, that if this number is 9, the following number must be 9 also, formally:
( d[n] = 9 ) => ( d[n+1] = 9 ) [I2]
Let's have a number S that equals 0.9999....9, and the number of nines past the decimal point is equal to n+1. Clearly, S < P. If our sought number Q have anything else on n+1 th position than 9, it's clear that immediately Q < S, and for every possible combination of the remaining digits in Q's decimal representation, Q will be always smaller than P, which is once again, contradiction, because for any digit on m-th position, where m>n, m-th digit in the representation of Q can never be higher than P's 9.
Now we use mathematical induction using [I1] (first number after decimal point must be 9) and [I2] (the next number must be always 9 too) to prove that our sought number Q must be expressed only by 0.9999...., and therefore is in fact equal to P.
P = Q, but that contradicts P < Q < 1, that means that we have proven by contradiction that no such real number Q that would fall between P and 1 exists. If no such real Q exists, then P must be equal to 1, i.e. 0.9999.... = 1
QED
Note - I don't know if this is entirely correct, I am out of school for some time.