What will be $((1 2 3 4)(1 5 7)^{-1}(2 4 3 6))^{111}$? Calculating the following permutation:
$$((1 2 3 4)(1 5 7)^{-1}(2 4 3 6))^{111}$$
Well, I can write that $(1 5 7)^{-1}=(7 5 1)$
So :$$((1 2 3 4)(7 5 1)(2 4 3 6))$$ which is equal to:
$$(1 2 3 4)^{111}(7 5 1)^{111}(2 4 3 6)^{111}$$
I can write the following:
$$111 \equiv -1 (mod 4) \implies 111 \equiv -1 (mod 3) \implies (1 2 3 4)^{111}=(1 2 3 4)^{-1}=(4 3 2 1)$$
then,
$$111 \equiv0( mod 3) \implies (1 2 3 4)^{0}=id $$
and 
$$ 111 \equiv -1( mod 4) \implies (2 4 3 6)^{111}=(2 4 3 6)^{-1}=(6 3 4 2)$$
so I get $$ (4 3 2 1)(6 3 4 2)$$
I'm not sure ,that my solution is correct, and how can I multiply $$ (4 3 2 1)(6 3 4 2)$$ in the very last step?
 A: $(1234)(751)(2436) = (1752)(36)(4)$ as disjoint cycles.
In that case powers behave nicely $((1752)(36))^{111} = (1752)^{111}(36)^{111}$
and $(36)^{k} = (36)$ for odd $k$, $(1)$ for even $k$.
And $111 \pmod {4} = -1$ so $(1752)^{111} = (1752)^{-1} = (2571)$.
So the answer to $((1234)(751)(2436))^{111} = (36)(2571)$
A: Observe that $(1234)(751)(2436)$ is the product of the three non-disjoint cycles: $$P = (2436),  \quad Q = (751) = (175), \quad R = (1234).$$  Then $RQP$ is found by noting that $P$ maps $1 \to 1$, $Q$ maps $1 \to 7$, then $R$ maps $7 \to 7$, so the cyclic decomposition begins with $(1 \, 7 \, \ldots)$.  Repeat the process so that $P : 7 \to 7$, $Q : 7 \to 5$, $R : 5 \to 5$, so the third term is $(1 \, 7 \, 5 \, \ldots)$.  We do this until we get back to $1$, so the first cycle in the decomposition is $(1752)$.  Then starting with $3$, the second cycle is $(36)$, and $4$ is fixed.
Now, to compute $((36)(1752))^{111}$, it is easy to see that disjoint cycles permute within themselves, so the composition of cycles distributes; $$((36)(1752))^{111} = (36)^{111} (1752)^{111}.$$  Since $(36)^2 = ()$ and $(1752)^4 = ()$, it follows that $(36)^{111} = 36$ and $(1752)^{111} = (1752)^3 = (1752)^{-1} = (1257)$.

Since you seem to think that Wolfram|Alpha gives a different result despite other answers agreeing with mine, I have provided Mathematica code to compute the cycle decomposition explicitly:
PermutationProduct[Cycles[{{2, 4, 3, 6}}], Cycles[{{1, 7, 5}}], Cycles[{{1, 2, 3, 4}}]]

yields
Cycles[{{1, 7, 5, 2}, {3, 6}}]

The next command computes the answer:
Nest[PermutationProduct[%, #] &, {}, 111]

which yields
Cycles[{{1, 2, 5, 7}, {3, 6}}].

It is important to note that when using PermutationProduct, the order in which the cycles are applied is from left-to-right as they appear in the list of arguments (so in this case, {2, 4, 3, 6} is applied first).  However, when the product of cycles is written in our usual cycle notation, the operation proceeds right-to-left.  This is why the order in which the cycles are presented is different.
A: As for your question on how to compute multiplications... I prefer to think using the two-line representation of permutations.
Take the example of $(1234)(751)(2436)$.  Begin writing the final result of the permutation by starting the first line as normal, having all elements in the correct order but leaving the bottom blank for now.
$$\begin{pmatrix}1&2&3&4&5&6&7\\\square&\square&\square&\square&\square&\square&\square\end{pmatrix}$$
We then try to fill in the blanks by seeing how the permutation as a whole acts on each element.  Having started with a specific element, if it is changed to a new element some time during the process, the next step we use that new element instead of the original.  Watching what happens to the element $1$, $(2436)$ will send $1\mapsto 1$.  Next $(751)$ maps $1\mapsto 7$.  Now, seeing how $(1234)$ acts, we see where it sends $7$ (because of the previous step) and we see it maps $7\mapsto 7$, so we can fill that piece of information in:
$$\begin{pmatrix}1&2&3&4&5&6&7\\7&\square&\square&\square&\square&\square&\square\end{pmatrix}$$
Similarly, we watch what happens to $2$:  $(2436)$ will map $2\mapsto 4$, $(751)$ will leave it alone, and $(1234)$ will map $4\mapsto 1$, so the permutation as a whole maps $2$ to $1$.
A bit faster now, $\begin{array}{}3\mapsto 6\mapsto 6\mapsto 6\\4\mapsto 3\mapsto 3\mapsto 4\\ 5\mapsto 5\mapsto 1\mapsto 2\\ 6\mapsto 2\mapsto 2\mapsto 3\\ 7\mapsto 7\mapsto 5\mapsto 5\end{array}$
So, we have our permutation is of the form: $$\begin{pmatrix}1&2&3&4&5&6&7\\7&1&6&4&2&3&5\end{pmatrix}$$
We may choose to write the above instead as a disjoint product of cycles, as that makes computing powers easier.  To convert a two-line notation permutation into a product of cycles, start a new cycle and start with $1$.  After that, write the number that $1$ is mapped to.  Write the number that that was mapped to, and again so on and so forth until we eventually reach $1$ again, instead of writing $1$ close the cycle.  Then begin a new cycle with the smallest remaining unused element and repeat the process.
Here, we begin by writing $1$ and the number below it:
$$(1~7\cdots$$
We continue, looking for the number below $7$ which was $5$.  That will be added to our cycle in progress.  Then the number below five, which is two, and the number below two which is again $1$
$$(1~7~5~2)(\cdots$$
The next smallest unused number would be $3$, and it gets mapped to $6$ and right back.  Finally $4$ just gets mapped to $4$.
$$(1~7~5~2)(3~6)(4)$$
Cycles with a single element in them are equal to the identity, so we may opt not to include $(4)$ in the decomposition above if we so choose.

Back to the original question, of $((1234)(751)(2436))^{111}$, as already mentioned in other answers, first change the representation of the permutation to be that of disjoint cycles so we may distribute the exponent (an operation which is otherwise not allowed).
This gets us $((1234)(751)(2436))^{111}=((1752)(36))^{111}=(1752)^{111}(36)^{111}=\dots$
Simplify the exponents on each cycle individually, and then perform multiplication once more on the final results.
