The smallest subgroup containing $ (1, 2) $ and $(1, 2, 3.,\ldots,n) $ is $ S_n$ I try  this by following way,
Let  $H $ be the subgroup  generated by $(1, 2), (1, 2, 3,\dots .,n) $. How do I show, $H$ contain  elements $ (1, r) $  for $ r = 1, 2,...n$. Does there exist any trick to show it? 
 A: We will show that every transposition can be generated from products of the two
permutations $(1, 2)$ and $(1, 2, . . . , n)$. This is achieved in $3$ separate steps
$1$. First observe,
$(1, 2, . . . , n)^{-1}(1, 2)(1, 2, . . . , n) = (2, 3)$
$(1, 2, . . . , n)^{-1}(2, 3)(1, 2, . . . , n) = (3, 4)$
$(1, 2, . . . , n)^{-1} (3, 4)(1, 2, . . . , n) = (4, 5)$
$(1, 2, . . . , n)^{-1}(n − 2, n − 1)(1, 2, . . . , n) = (n − 1, n)$
In short, transpositions of the form $(i, i + 1)$ have been obtained.
$2$. Thus,
$(1, 2)(2, 3)(1, 2) = (1, 3)$
$(1, 3)(3, 4)(1, 3) = (1, 4)$
$(1, 4)(4, 5)(1, 4) = (1, 5)$
.
.
.
$(1, n − 1)(n − 1, n)(1, n − 1) = (1, n)$
In short, transpositions of the form $(1, i)$ have been obtained.
$3$. Finally, $(1, i)(1, j)(i, 1) = (i, j)$ for$ i \neq j$
Hence, every transposition has been generated from $(1, 2)$ and $(1, 2, . . . , n)$, so these generate all of $S_n$
A: Let $s = (1\;2)$ and $c = (1\;2\;3\;\ldots\;n)$ then $$c^{-i} s c^{i} = (i\;i+1)$$ and since you can create any permutation from transpositions, this gives the whole of $S_n$.
To verify this identity, see that $j$ gets mapped to $j-i \mod n$ then swapped only if it's 1 or 2, then $i$ is added back.

Also $c^{-(n-2)} s c^{n-1} = (1\;2\;3\;\ldots\;n-1)$

If we use $a^b$ short for $b^{-1} a b$ (the reason this is such a useful operation to do is that it preserves the length of the cycle) then consider


*

*$s$

*$(s^c)^s$

*$(((s^c)^s)^c)^s$

*...

