Seeking some easier method to prove the identity

Question

I came across the identity

Let $n$ and $m$ be natural numbers and $p_{1}, p_{2}, p_{3}...p_{n}$ be real numbers. Prove that if any $a_{1},a_{2},..a_{k}$ for any $ k\le n$ ramdomly chosen from the set $S$ = {$p_{1}, p_{2}, p_{3}...p_{n}$} follow $$ \prod_{i=1}^{k} a_{i} = \sum_{i = 1}^{k}a_{i}-k+1$$ then $$(\prod_{i=1}^{n} p_{i})^{m} = (\sum_{i = 1}^{n}p_{i}-n+1)^{m} = \sum_{i=1}^{n}p_{i}^{m}-n+1 $$

MY TRY :- For e.g. for $n=2$ we have $pq=p+q-1$ and we need to prove that $(p+q-1)^{m} = p^{m}+q^{m}-1$ but as pointed out by Wolfgang Kais it is fairly simple as the equation $p+q-1 = pq$ is satisfied if and only if $(p-1)(q-1) = 0\implies$ either $p=1$ or $q=1$ which then is simple to prove.

I also proved it for any n using induction. I have written the method in answer. But it took a large calculation to do it. I suppose there will definitely be some easier method to prove it.
Any help will be highly appreciated.
Please ask for clarifications in case of discrepencies.

Answer 1

Knowing the prerequisit that was previously missing, the proof can indeed be simplified. To be more specific, we can conclude that all but at most one of the numbers are equal to $1$. We can use the case $n=2$ to do so.

For $n=1$, we can choose any real number $p_1$ and trivially at most one of this one number differs from $1$.

Now, let $n\ge2$ be an integer and $p_1,...,p_n$ be real numbers such that for any integer $k \le n$ and any $k$-subset $\{j_1, ..., j_k\} \subseteq \{1,...,n\}$ we have $$ \prod_{i=1}^k p_{j_i}=\sum_{i=1}^k p_{j_i}-(k-1) $$ Then, for any pair $(i,j)$ with $i,j \in \{1,...,n\}$, $i \ne j$, we have $$\begin{align} p_i \cdot p_j &= p_i+p_j-(2-1)\\ \Leftrightarrow 1-p_i-p_j+p_i\cdot p_j&=0\\ \Leftrightarrow (1-p_i)\cdot (1-p_j)&=0\\ \Leftrightarrow p_i=1 \lor p_j=1& \end{align}$$ Therefore, at least one member of every pair of the numbers is equal to $1$. This can only be true if at most one of the numbers differs from $1$. So, all but at most one of the numbers are equal to $1$. W.l.o.g, we assume that $p_1=...=p_{n-1}=1$ and we conclude for any integer $m \ge 1$: $$ (\prod_{i=1}^n p_i)^m=p_n^m = p_n^m+\sum_{i=1}^{n-1}1^m-(n-1)=\sum_{i=1}^{n}p_i^m-(n-1) $$

Answer 2

We'll use 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 $$ =\sum_{j=1}^{n}p_{j}^{k+1}-t\Bigl(\sum_{i=1}^{n}p_{i}-\prod_{i=1}^{n}p_{i}\Bigr)+t^2-\sum_{j=1}^{n}p_{j}^{k} +\biggl(\sum_{j=1}^{n}p_{j}-t\biggr)^{k}$$ $$ =\sum_{j=1}^{n}p_{j}^{k+1}-t\Bigl(t\Bigr)+t^2-\sum_{j=1}^{n}p_{j}^{k} +\Bigl(\sum_{j=1}^{n}p_{j}^{k}-t\Bigr) = \sum_{j=1}^{n}p_{j}^{k+1}-t.$$