Proving $S_n$ is generated by the set $\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$ of $n-1$ transpositions.

I am struggling with the end of my proof.

So far I have used the fact that $S_n=\langle T\rangle$ where $T=\{(i\text{ }j)|1\leq i<j\leq n\}$ to show that $S_n$ is generated by the set $\{(1\text{ }2),(1\text{ }3),\cdots,(1\text{ }n)\}$ of $n-1$ transpositions.

I am struggling with proving that $S_n$ is also generated by the set $\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$ of $n-1$ transpositions. Essentially I want to show that each $(1\text{ }i)$ is a product of these transpositions to reach this conclusion.

Here is what I have so far...

Consider any $(1\text{ }i)$ transposition in $S_n$. Notice by the composition of a permutation that \begin{eqnarray*} (1\text{ }i-1)(i-1\text{ }i)(1\text{ }i-1)&=&\begin{bmatrix} 1 & i-1 & i \\ i-1 & 1 & i \end{bmatrix} \begin{bmatrix} 1 & i-1 & i \\ 1 & i & i-1 \end{bmatrix} \begin{bmatrix} 1 & i-1 & i \\ i-1 & 1 & i \end{bmatrix}\\ &=& \begin{bmatrix} 1 & i-1 & i \\ i-1 & 1 & i \end{bmatrix} \begin{bmatrix} 1 & i-1 & i \\ i & 1 & i-1 \end{bmatrix}\\ &=& \begin{bmatrix} 1 & i-1 & i \\ i & i-1 & 1 \end{bmatrix}\\ &=& (1\text{ }i) \end{eqnarray*}

Now we want to use this property, call it STAR, to show that $S_n$ is also generated by the set $$Y=\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$$ of $n-1$ transpositions. Notice for each $(i\text{ }j)$ in $Y$ that $j-i=1$. Suppose $n=2$. Then $S_2=\{(1),(1\text{ }2)\}$, and we done because $2-1=1$. Suppose $n=3$. Then $S_3$ is generated by the set $\{(1\text{ }2),(1\text{ }3)\}$. By STAR, $$ (1\text{ }3)=(1\text{ }2)(2\text{ }3)(1\text{ }2) $$ $$ (1\text{ }2)=(1\text{ }1)(1\text{ }2)(1\text{ }1) = (1\text{ }2) $$ Therefore $S_3$ is generated by $\{(1\text{ }2),(2\text{ }3)\}$ and for each $(i\text{ }j)$ we have $j-i=1$.

And here is where I am struggling. I do not know how to apply induction to this. I want to show that each side part will continue breaking down using the property I stated until all of $S_n$ is created. Here is why I think this: $$ (1\text{ }1)(1\text{ }2)(1\text{ }1)=(1\text{ }2) $$ $$ (1\text{ }2)(2\text{ }3)(1\text{ }2)=(1\text{ }3) $$ $$ \cdots $$ $$ (1\text{ }n-1)(n-1\text{ } n)(1\text{ } n-1)=(1\text{ }n) $$ Any input would be much appreciated. Best.

There is a very detailed section about generating sets for $S_n$ in K. Conrad's notes. Theorem $2.3$ on page $3$ proves that the transpositions $\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$ of $n-1$ generate $S_n$. It reduces to permutations of type $(i,i+1)$, as you said, and uses that $S_n$ is generated by all transpositions, too (which is easy). The proof is so well-written, that I will not try to repeat it. I recommend very much to read this section there, to see whether your induction is correct.