I was interested if this equality holds for arbitrary $n$:
$$(1 \space 2)(2\space3)(3\space 4)...(n-1\space n) (n\space 1)=(1)(2 \space 3 \space 4... \space n-1 \space n)$$
(considering multiplications are done in $S_n$)
It looks like it is true (I've done calculations for $n=2, 3, 4$ and the formula does work). But it's hard for me to prove this in general.
We assume that permutations act from the left, so we apply the transpositions from right-to-left.
As the OP mentions in the comments, $1$ is a fixed point of the permutation in question.
The element $n$ follows the trajectory $n \mapsto 1 \mapsto 2$.
For any other $k$ with $1 < k < n$, element $k$ is untouched until it encounters the transposition $(k\;k+1)$, at which point it moves to $k+1$ which puts it out of reach of all other transpositions.
Hence $k \mapsto k+1$ for all such $k$. Combining this with $n \mapsto 2$ gives the desired cycle structure.