"pi" here just represents an arbitrary permutation, it's just a "variable symbol" (like x, or y, or n), and does not refer to a SPECIFIC permutation (it's just often used as a "typical permutation" because "pi" is the greek letter corresponding to "p" for "permutation").
let me elaborate on what Mr. Pinter is getting at:
a given permutation, like say (1 2 3) in S3, can be written as a product of transpositions in more than one way:
(1 2 3) = (1 3)(1 2)
(1 2 3) = (2 3)(1 3)
(1 2 3) = (1 2)(1 3)(1 2)(1 3)
or, even more trivially:
(1 2 3) = (1 3)(1 2)(1 2)(1 2)
note that all of these have an EVEN number of transpositions in the product (2 for the first 2, 4 for the second 2).
but...how do we know there is NO way to write (1 2 3) as a product of 7 transpositions, or 5, or 23? that is, how do we know the PARITY of the number of transposition factors is invariant? (parity is just a fancy word for "even- or odd-ness").
this is not so obvious...there are clearly an infinite number of ways to write any permutation as a product of 2-cycles (we can always insert a pair (a b)(a b) in-between any such product to get a new one equal to the old one), so a direct proof is not going to be easy.
the way this is normally proved is to use a very special polynomial in n variables:
it can be shown that for a permutation σ in Sn:
in particular, if σ is a transposition, then:
so if a permutation σ fixes the sign of p, it must change the sign of an even number of the factors of p, and for a permutation to be both even AND odd, we would have to have p = -p, and thus p = 0, which is clearly not the case.
i urge you to play around with this for the case n = 3, in which case p is the polynomial:
(x1 - x2)(x1 - x3)(x2 - x3) to see how this actually works.