Solving $p_1^{e_1} p_2^{e_2}...p_k^{e_k}=e_1^{p_1} e_2^{p_2}...e_k^{p_k}$ Find all positive integers $k$, positive integers $e_i$, and distinct prime numbers $p_i$ for $1\le i\le k$, such that $$p_1^{e_1} p_2^{e_2}...p_k^{e_k}=e_1^{p_1} e_2^{p_2}...e_k^{p_k}.$$
Is this impossible? I've been trying to find solutions for at least a week. I've made a conjecture but I wish to suppress it for the time being (I don't think it will be of use to anyone attempting this).
Thanks for any help.
 A: I am only answering this to draw attention to the correct solution set to this problem; I am not intending to accept my own answer.


*

*If the $e_i$ are primes then the only solutions are of the form $e_i=p_{\sigma(i)}$ for any permutation of the set $\{1,2,...,k\}$ with the property that it contains only cycles of length 1 and/or 2.

*If not all the $e_i$ are primes, then exactly $k-1$ of them are prime (of the form in (1)) and the remaining $p_m^{e_m}$, $e_m^{p_m}$ pair satisfies $(p_m,e_m)=(2,4)$.
If anyone can prove this, they are a genius.
A: EDIT I have edited the lemma's demonstration to make it directly applicable to the problem and corrected some typos in the formulas.
Lemma For any $k$ integers $(\alpha_i)_{1\leq i\leq k}$ and distinct $(n_i)_{1\leq i\leq k}$ such that for all $i$, $\alpha_i\geq0$ and $n_i\geq2$, we have
$$  \prod_{i=1}^k n_i^{\alpha_i}\geq\sum_{i=1}^k\alpha_i n_i\tag 1$$
with equality if and only if exactly one of the $\alpha_i$ is nonzero 
and $\alpha_i=1$ or $n_i=\alpha_i=2$.
Proof: 
We first prove that for $n\geq2$ and $\alpha\geq0$, $\alpha n\leq n^\alpha$
and give the equality cases.


*

*The case $\alpha=0$ is a strict inequality $0<1$.

*The case $\alpha=1$ is an equality case. 

*In the case $\alpha=2$, with $n=2$ we have another equality case $2^2=2\times2$.

*For $\alpha=3$ and $n=2$, we have the strict inequality $n^3=8>6=3n$. For $n\geq3$, $n^2\geq3n>2n$ so we have the strict inequality for $\alpha=2$. Suppose now that for $n\geq2$ we have for some $\alpha\geq2$ the inequality  $n^\alpha>\alpha n$. We have therefore $n^\alpha+n>\alpha n+n$ or 
$$n\times\left(n^{\alpha-1}+1\right)>(\alpha+1)n.$$
Clearly $n^\alpha\geq n^{\alpha-1}+1$ because $\alpha-1\geq 1$. We conclude that $n^{\alpha+1}>(\alpha+1)n$ which proves the required result by recurrence on $\alpha$.


In the next step, we consider $k$ distinct integers larger than $2$. Then the sum of the integers is equal to their product if and only if $k=1$. For two integers $x\geq2$ and $y\geq2$ we have indeed $xy\geq2\max\{x,\,y\}>x+y.$
Consider now the numbers $n_i^{\alpha_i}$ with $\alpha_i>0$. We can apply the product inequality and get 
$$\prod_{\substack{i\\\alpha_i\neq0}}n_i^{\alpha_i}\geq
\sum_{\substack{i\\\alpha_i\neq0}}n_i^{\alpha_i}\geq\sum_{\substack{i\\\alpha_i\neq0}}\alpha_in_i.$$
The first inequality is an equality if and only if only one $\alpha_i\neq0$.
The second inequality is an equality if and only if this $\alpha_i=1$ or $\alpha_i=n_i=2$.
We can remove the statement $\alpha_i\neq0$ to get the final inequality (1) because factors with $\alpha_i=0$ in the left-hand side product are equal to $1$ and terms with $\alpha_i=$ in the right-hand side sum are equal to $0$.  
Proof of the main result
Consider an equality as written by the OP. We can write the number $e_i$ as
$$ e_i=p_1^{\alpha_{i,1}}\dots\,p_k^{\alpha_{i,k}}$$ with $\alpha_{i,j}\geq0$ because no other prime factor is possible for $e_i$'s than the $p_i$'s. Let us replace in the equality. We get
$$ p_1^{p_1^{\alpha_{1,1}}\dots\,p_k^{\alpha_1,k}}\dots\, p_k^{p_k^{\alpha_{k,1}}\dots\,p_k^{\alpha_k,k}}
=p_1^{\alpha_{1,1}p_1+\cdots+\alpha_{k,1}p_k}\dots\,
p_k^{\alpha_{1,k}p_1+\cdots+\alpha_{k,k}p_k}.$$
Let us identify the exponents of $p_i$ on both sides. We get
$$ e_i=\prod_{j=1}^kp_j^{\alpha_{i,j}}=\sum_{j=1}^k\alpha_{j,i}p_j.$$
Using the lemma lemma with the left-hand side product we get the inequality
$$e_i=\prod_{j=1}^kp_j^{\alpha_{i,j}}\geq\sum_{j=1}^k\alpha_{i,j}p_j\tag{2}$$
and we conclude that for all $i$
$$ d_i=\sum_{j=1}^k\left(\alpha_{i,j}-\alpha_{j,i}\right)p_j\geq0.$$
Let us compute the sum
$$S=\sum_{i=1}^kp_id_i=\sum_{1\leq i,j\leq k}\left(\alpha_{i,j}-\alpha_{j,i}\right)p_ip_j\geq0,$$
we can swap the indices in the sum and get
$$S=\sum_{1\leq i,j\leq k}\left(\alpha_{j,i}-\alpha_{i,j}\right)p_jp_i=-S.$$
As a conclusion, we have $S=0$. Since $d_i\geq0$, we must have $d_i=0$ for all $i$ and consequently $\alpha_{i,j}=\alpha_{j,i}$ for all $i,j$.
Therefore the inequality in (2) is in the equality case. We conclude from the lemma that for any given $i$, there is a unique $j$ such that $e_i=p_j^{\alpha_{i,j}}=\alpha_{i,j}p_j$ and that $\alpha_{i,j}=1$ (then $e_i=p_j$) or $\alpha_{i,j}=2$ (then $e_i=4$, $p_j=2$). The latter case implies $p_j=2$ so, as the $p_i$'s are distinct, there can be only one $\alpha_{i,j}$ equal to $2$, so $i=j$.
Excluding such a pair ($p_i=2$, $e_i=4$), we obtain that 
there is a permutation $\sigma$ of $\{1,\dots\,k\}$ satisfying $e_i=p_{\sigma(i)}$ for all other $i$. Moreover, the identity $\alpha_{i,j}=\alpha_{j,i}$ implies that $\sigma^2$ is the identity, that is $\sigma$ has no cycle longer than $2$. 
As a conclusion, all solutions are given by a set of $k$ prime numbers and one the following 


*

*a permutation $\sigma$ of $\{1,\,\dots,\,k\}$ with no cycle longer than $2$. Then $e_i=p_{\sigma(i)}$ is a solution.

*if $2$ is among the primes, say $p_1=2$, a solution with the $k-1$ other prime numbers for $e_2,\,\dots e_k$ and $e_1=4$.
