Prove that for disjoint permutations $P_{1},P_{2}..P_{n}$ we have $(\prod_{i=1}^{n} P_{i})^{m} = \sum_{i=1}^{n}P_{i}^{m}-(n-1)I $ 
Let $n$ and $m$ be natural numbers and $P_{1}, P_{2},...P_{n}$ be permutation matrices which are represent pairwise disjoint permutations then
Prove that  $$\biggl(\prod_{i=1}^{n} P_{i}\biggr)^{m} =  \biggl(\sum_{i = 1}^{n}P_{i}-(n-1)I\biggl)^{m} = \sum_{i=1}^{n}P_{i}^{m}-(n-1)I $$

Before we start let's have the following result in hand for any number of  pairwise disjoint matrices we have  $$ \prod_{i=1}^{n} P_{i} = \sum_{i = 1}^{n}P_{i}-(n-1)I$$ which is proved here.
As disjoint matrices are commutative, while multiplying them we can use them as numbers $p_{1}, p_{2},...p_{n}$
I used Induction to prove it. But it was much calculative. I have written my try in the answer. I am curious to see if some relatively simpler method to prove this exists? Please ask for clarifications in case of any discrepencies. Any hint will be a great help!
 A: I assume that if $A$ is a matrix and $k$ is a scalar, then $A + k$ is meant to denote $A + kI$, where $I$ denotes the identity matrix.
Because the $P_i$ represent pairwise disjoint permutations, note that $(P_i - 1)(P_j - 1) = 0$ whenever $i \neq j$.
Now, take $\left(\prod_{i=1}^n P_i \right)^m$ and expand the product within to get
$$
\left(\prod_{i=1}^n P_i \right)^m = 
\left(\prod_{i=1}^n [1 + (P_i - 1)] \right)^m = 
\left(1 + \sum_{i=1}^n (P_i - 1)\right)^m.
$$
Simplify the sum inside to get $\left(1 + \sum_{i=1}^n (P_i - 1)\right)^m = \left(-(n-1) + \sum_{i = 1}^{n}P_{i}\right)^{m}$. This gives us one equality.
From there, expand the $m$th power to get
$$
\left(1 + \sum_{i=1}^n (P_i - 1)\right)^m = 1 + \sum_{i=1}^m\sum_{j=1}^n \binom mi (P_j - 1)^i. 
$$
By the binomial theorem, we can rewrite the right hand side of the above as
$$
1 + \sum_{i=1}^m\sum_{j=1}^n \binom mi (P_j - 1)^i = 1 + \sum_{j=1}^n \left[(1 + (P_j - 1))^m - 1\right] = -(n-1) + \sum_{j=1}^nP_j^m.
$$

Alternatively, we could prove $-(n-1) + \biggl(\prod_{i=1}^{n} P_{i}\biggr)^{m} = \sum_{i=1}^{n}P_{i}^{m}$ as follows.
Note that because the permutations $P_i$ commute, we have
$$
\biggl(\prod_{i=1}^{n} P_{i}\biggr)^{m}= \prod_{i=1}^n P_i^m.
$$
Now, the permutation matrices $P_1^m, \dots, P_n^m$ represent pairwise disjoint permutations. Using either the "answer in hand" you linked or my work above, it follows that
$$
\biggl(\prod_{i=1}^{n} P_{i}\biggr)^{m}= \prod_{i=1}^n P_i^m = -(n - 1) + \sum_{i=1}^n P_i^m,
$$
which was what we wanted.
A: I used strong form of induction on $m$, let $n$ be arbitrary, base case $m = 1$ is trivial, let   $$ \Bigl(\sum_{i=1}^{n}p_{i}-n+1\Bigr)^{m} = \sum_{i=1}^{n}p_{i}^{m}-n+1    \space \forall\space 2\le m\leq k$$ we'll prove that it also holds for $m = k+1$. Let $n-1 = t$ to reduce the congestion.
$$\biggl(\sum_{i=1}^{n}p_{i}-t\biggr)^{k+1} = \biggl(\sum_{j=1}^{n}p_{j}-t\biggr)^{k}\biggl(\sum_{i=1}^{n}p_{i}-t\biggr) = \biggl(\sum_{j=1}^{n}p_{j}^{k}-t\biggr)\biggl(\sum_{i=1}^{n}p_{i}-t\biggr)$$ $$   
 = \sum_{i=1}^{n}-t(p_{i}+p_{i}^{k})+t^{2}+\sum_{j=1}^{n}p_{j}^{k}\sum_{i=1}^{n}p_{i}$$Call the above equation [$1$]. The last term in [$1$]$$\sum_{j=1}^{n}p_{j}^{k}\sum_{i=1}^{n}p_{i} = \sum_{j=1}^{n}p_{j}^{k}\sum_{i\neq j}^{n}p_{i}+ \sum_{j=1}^{n}p_{j}^{k+1} = \sum_{j=1}^{n}p_{j}^{k}\biggl(\prod_{i\neq j}^{n}p_{i}+n-2\biggr)+\sum_{j=1}^{n}p_{j}^{k+1}$$
$$= \sum_{j=1}^{n}p_{j}^{k-1}\prod_{i=1}^{n}p_{i}+(n-2)\sum_{j=1}^{n}p_{j}^{k}+\sum_{j=1}^{n}p_{j}^{k+1}$$
Call this equation [$2$]. First term in [2]$$\sum_{j=1}^{n}p_{j}^{k-1}\prod_{i=1}^{n}p_{i}=\prod_{i=1}^{n}p_{i}\biggl[\sum_{j=1}^{n}p_{j}^{k-1}-t\biggr]+t\prod_{i=1}^{n}p_{i}$$
$$=\prod_{i=1}^{n}p_{i}\biggl(\sum_{j=1}^{n}p_{j}-t\biggr)^{k-1}+t\prod_{i=1}^{n}p_{i}=\biggl(\sum_{j=1}^{n}p_{j}-t\biggr)^{k}+t\prod_{i=1}^{n}p_{i}$$
Coming back to [$1$] RHS
$$ = -t\sum_{i=1}^{n}p_{i}-t\sum_{i=1}^{n}p_{i}^{k}+t^{2}+\Biggl(\biggl(\sum_{j=1}^{n}p_{j}-t\biggr)^{k}+t\prod_{i=1}^{n}p_{i}+(t-1)\sum_{j=1}^{n}p_{j}^{k}+\sum_{j=1}^{n}p_{j}^{k+1}\Biggr)$$
Arranging the terms $$ =\sum_{j=1}^{n}p_{j}^{k+1}-t\Bigl(\sum_{i=1}^{n}p_{i}-\prod_{i=1}^{n}p_{i}\Bigr)+t^2-\sum_{j=1}^{n}p_{j}^{k} +\biggl(\sum_{j=1}^{n}p_{j}-t\biggr)^{k}$$
$$ =\sum_{j=1}^{n}p_{j}^{k+1}-t\Bigl(t\Bigr)+t^2-\sum_{j=1}^{n}p_{j}^{k} +\Bigl(\sum_{j=1}^{n}p_{j}^{k}-t\Bigr) = \sum_{j=1}^{n}p_{j}^{k+1}-t.$$
