Ytukyg

Prove: if $sigmain S_n$ is a factorization of $c$ disjoint cycles then $text{sgn} (sigma)=(-1)^{n-c}$.

We know the one cycle sign is $(-1)^{l-1}$ so $c$ of them is:

$$text{sgn}(sigma)=(-1)^{l-1}cdot (-1)^{l-1}cdotcdotcdot(-1)^{l-1}=(-1)^{ccdot (l-1)}=(-1)^{ccdot l-c}$$
and because they are disjoint cycles $n=ccdot l$ and so $text{sgn}(sigma)=(-1)^{n-c}$.

Is this proof valid?

According to the proven sentence, the length of each cycle can be different, in Wikipedia there is the following fact:

"In practice, in order to determine whether a given permutation is even or odd, one writes the permutation as a product of disjoint cycles. The permutation is odd if and only if this factorization contains an odd number of even-length cycles."

Is this another way to calculate the sign? Or it is a special case of the sentence?

Your proof as it stands only works if there are $c$ disjoint cycles of an equal length, for example $(13)(26)(45)=(362542)$.

A more formal proof can be written as:

Prove: if $sigmain S_n$ is a factorization of $c$ disjoint cycles
with $o$ odd length cycles and $e$ even length cycles, then
$text{sgn}(sigma)=(−1)^{n−c}=(−1)^e$.

Odd length cycles always have sign $1$. After removing them, we have a
permutation on $n-o$ elements, and this must be even as we only have
even cycles left. So $nequiv o; text{mod}2$ and we can write:
$$(-1)^{n-c}=(-1)^{n-o-e}=(-1)^{n-o}cdot (-1)^{-e}=1cdotdfrac{1}{(-1)^e}=(-1)^e$$

Your argument can be rescued here a bit, the parity of an even length
cycle is always $-1$, so let $l=2k$ in your argument, and we can argue
that:

We know the cycle sign for a cycle of even length is $(-1)^{2k-1}=-1$
so $e$ of them equals $(-1)^e$ and so $text{sgn}(sigma)=(-1)^e=(-1)^{n-c}$.

First, a matter of terminology: if, like most authors, you consider cycles to be of length $>1$, then your statement is false. In order for it to be true, we shall agree to also speak of cycles of length $1$ (i.e. the fixed points of permutations).

If $sigma_1, sigma_2$ are disjoint cycles of lengths $l_1, l_2$, then $sigma_1 sigma_2$ acts on precisely $l_1 + l_2$ elements (the ones permuted by $l_1$ and $l_2$), leaving the other $n - l_1 - l_2$ fixed. If $sigma = sigma_1 sigma_2 dots sigma_c$ then, by induction, $sigma$ acts on precisely $l_1 + l_2 + dots + l_c$ elements, and this number is precisely $n$ (because a decomposition into disjoint cycles exhausts a permutation). Now, $$mathrm{sgn} ; sigma = mathrm{sgn} ; sigma_1 mathrm{sgn} ; sigma_2 dots mathrm{sgn} ; sigma_c = (-1)^{l_1 -1} (-1)^{l_2 -1} dots (-1)^{l_c -1} = (-1)^{l_1 + l_2 dots l_c -c} = (-1)^{n-c} ,$$ which is your desired result.