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Seems to be really addictive, so you will have to endure one more answer I am afraid.

What follows is actually present in several of already given answers, I am just trying to make it as simple as possible.

Write down $\binom n2$ statements "$1<2$", "$1<3$", ..., "$n-1<n$".

Now perform a permutation and count how many of these statements will become violated.

If this permutation is a transposition $(ij)$, for $i<j$, then those violated are all "$i<k$" with $k<j$, all "$\ell<j$" with $\ell>i$ (same number twice), and "$i<j$". So each transposition violates an odd number of these statements.

Viewing result of a permutation as a reordering, we see that performing one more transposition switches between "even violators" and "odd violators".

I believe this suffices to convince oneself that parity of the number of transpositions producing any given permutation (unlike the number itself) is unambiguously defined, and coincides with the parity of the number of those violations.

Seems to be really addictive, so you will have to endure one more answer I am afraid.

What follows is actually present in several of already given answers, I am just trying to make it as simple as possible.

Write down $\binom n2$ statements "$1<2$", "$1<3$", ..., "$n-1<n$".

Now perform a permutation and count how many of these statements will become violated.

If this permutation is a transposition $(ij)$, for $i<j$, then those violated are all "$i<k$" with $k<j$, all "$\ell<j$" with $\ell>i$ (same number twice), and "$i<j$". So each transposition violates an odd number of these statements.

Viewing result of a permutation as a reordering, we see that performing one more transposition switches between "even violators" and "odd violators".

I believe this suffices to convince oneself that parity of the number of transpositions producing any given permutation (unlike the number itself) is unambiguously defined, and coincides with parity of the number of those violations.

Seems to be really addictive, so you will have to endure one more answer I am afraid.

What follows is actually present in several of already given answers, I am just trying to make it as simple as possible.

Write down $\binom n2$ statements "$1<2$", "$1<3$", ..., "$n-1<n$".

Now perform a permutation and count how many of these statements will become violated.

If this permutation is a transposition $(ij)$, for $i<j$, then those violated are all "$i<k$" with $k<j$, all "$\ell<j$" with $\ell>i$ (same number twice), and "$i<j$". So each transposition violates an odd number of these statements.

Viewing result of a permutation as a reordering, we see that performing one more transposition switches between "even violators" and "odd violators".

I believe this suffices to convince oneself that parity of the number of transpositions producing any given permutation (unlike the number itself) is unambiguously defined, and coincides with the number of those violations.

Seems to be really addictive, so you will have to endure one more answer I am afraid.

What follows is actually present in several of already given answers, I am just trying to make it as simple as possible.

Write down $\binom n2$ statements "$1<2$", "$1<3$", ..., "$n-1<n$".

Now perform a permutation and count how many of these statements will become violated.

If this permutation is a transposition $(ij)$, for $i<j$, then those violated are all "$i<k$" with $k<j$, all "$\ell<j$" with $\ell>i$ (same number twice), and "$i<j$". So each transposition violates an odd number of these statements.

Viewing result of a permutation as a reordering, we see that performing one more transposition switches between "even violators" and "odd violators".

I believe this suffices to convince oneself that parity of the number of transpositions producing any given permutation (unlike the number itself) is unambiguously defined, and coincides with the parity of the number of those violations.

Seems to be really addictive, so you will have to endure one more answer I am afraid.

What follows is actually present in several of already given answers, I am just trying to make it as simple as possible.

Write down $\binom n2$ statements "$1<2$", "$1<3$", ..., "$n-1<n$".

Now perform a permutation and count how many of these statements will become violated.

If this permutation is a transposition $(ij)$, then those violated are all "$i<k$" with $k<j$, all "$\ell<j$" with $\ell>i$ (same number twice), and $i<j$. So each transposition violates an odd number of these statements.

Viewing result of a permutation as a reordering, we see that performing one more transposition switches between "even violators" and "odd violators".

Seems to be really addictive, so you will have to endure one more answer I am afraid.

What follows is actually present in several of already given answers, I am just trying to make it as simple as possible.

Write down $\binom n2$ statements "$1<2$", "$1<3$", ..., "$n-1<n$".

Now perform a permutation and count how many of these statements will become violated.

If this permutation is a transposition $(ij)$, for $i<j$, then those violated are all "$i<k$" with $k<j$, all "$\ell<j$" with $\ell>i$ (same number twice), and "$i<j$". So each transposition violates an odd number of these statements.

Viewing result of a permutation as a reordering, we see that performing one more transposition switches between "even violators" and "odd violators".

I believe this suffices to convince oneself that parity of the number of transpositions producing any given permutation (unlike the number itself) is unambiguously defined, and coincides with the number of those violations.