Your solution is basically correct, though not very well formulated. But you are not clear about how you represent a permutation, and whether "applying" a permutation corresponds to multiplication on the left or on the right, and this makes it difficult to point precisely to what you should do different. But in any case you can salvage your kind of reasoning by making the proper choices.
In group theory permutations are defined as functions (bijections), not as permuted sequences. So first let us fix the correspondence: a sequence $(a_1,\ldots,a_n)$ (with $\{a_1,\ldots,a_n\}=\{1,\ldots,n\}$) represents the permutation $\sigma$ that sends $i\mapsto a_i$ for all$~i$. (This seems a natural choice, but a choice nonetheless.) Next we need to be clear about left and right in permutation composition; since you write (permutation) functions before their argument, let us choose as usual that the composition $f.g$ is $i\mapsto f(g(i))$ (the permutation on the right gets to operate first). Again this is a natural choice, but some prefer to have $f$ act first (typically those who prefer to write $i^f$ instead of $f(i)$); I just want to make my choice clear.
With these choices, if $\sigma$ is represented by $(a_1,\ldots,a_n)$, then $\pi.\sigma$ is represented by
$$
\bigl(\pi(\sigma(1)),\ldots,\pi(\sigma(n))\bigr)
= \bigl(\pi(a_1),\ldots,\pi(a_n)\bigr),
$$
in other words left multiplication by$~\pi$ corresponds to applying the function $\pi$ to the values of individual entries of the sequence, not to permuting the entries of the sequence according to$~\pi$. However if we multiply on the right by$~\pi$, the situation is different: $\sigma.\pi$ is represented by
$$
\bigl(\sigma(\pi(1)),\ldots,\sigma(\pi(n))\bigr)
= \bigl(a_{\pi(1)},\ldots,a_{\pi(n)}\bigr).
$$
Here it is the positions that are permuted. Since you use this in your argument, it can be made valid if you stipulate that each successive permutation applied to the sequence corresponds to a right-multiplication.
I should warn about a slight twist that remains: right multiplication by $\pi$ does permute the entries in the sequence, but it does so in such a way that the new entry at position $i$ is $a_{\pi(i)}$, which is the one that used to be in position$~\pi(i)$; the old value $a_i$ will be found after permutation at position $\pi^{-1}(i)$. This is what you probably associate with permuting the entries according to the inverse permutation$~\pi^{-1}$. Indeed this is true for the usual definition of permutations acting (from the left!) on sequences, as I detailed in this answer. If you think of it, it is normal that when right-multiplication is related to a left-action on sequences, an inversion should be used in the process.
No doubt this somewhat unfortunate consequence of the otherwise natural choices above is what motivates people to make the opposite choice for one of the two. As I said it is often the second that is reversed, but this has other notational consequences that can be confusing. For me the real culprit is the first choice: it would be more natural to represent $\sigma$ as the result of permuting the standard sequence $(1,2,\ldots,n)$ according to$~\sigma$, and this gives $(\sigma^{-1}(1),\sigma^{-1}(2),\ldots,\sigma^{-1}(n))$ rather than $(\sigma(1),\sigma(2),\ldots,\sigma(n))$. However, I would not really advocate doing this, as this would make the confusion between notations even worse.
Yes. Start with the permutation which lists the numbers $1$ through $N$ in increasing order. Then repeatedly do the following: In the permutation you have, find the last pair of consecutive numbers in which the latter is greater than the former (if there is no such pair, stop, you're done). Call this pair $mn$. Increase $m$ by $1$ until you get a number $m'$ that does not appear prior to $m$. Then list the remaining (not yet listed) numbers after $m'$ in increasing order.
Example:
1234
1243
1324
1342
1423
1432
2134
2143
2314
2341
2413
2431
3124
3142
3214
3241
3412
3421
4123
4132
4213
4231
4312
4321
Best Answer
Yes! You are asking if the cayley graph of the symmetric group with a basis of transpositions is hamiltonian.
In fact this is true, and was proven in Kompel'makher and Liskovet's "Sequential Generation of Arrangements by Means of a Basis of Transpositions".
This is only a 5-page paper, and a fairly easy read. Moreover, it describes an algorithm for finding these cycles using only the swaps $(1,i)$ for $2 \leq i \leq n$, the so called "star-shaped basis".
I hope this helps ^_^