TITLE: Prime exponent Diophantine equations with infinitely many solutions
QUESTION [5 upvotes]: I am to prove that the following equation has infinitely many solutions in the set of natural numbers:
$${x_1}^{p_1} + {x_2}^{p_2} + \cdots + {x_{n-1}}^{p_{n-1}} = {x_n}^{p_n},$$ where $p_1, \dots ,p_n$ are given different prime numbers. 
Should one resort to induction, checking the first step is not so difficult, as the case for $n=1$ is obvious, and as to $n=2$ for the equation ${x_1}^{p_1} = {x_2}^{p_2}$ one could set $x_1= p^{p_2} \, , x_2=p^{p_1}$ for an arbitrary prime $p$. 
But in passing from the induction hypothesis (the case $n$) to the case $n+1$, I could not get through. Maybe, of course, the problem is solved without induction. Thanks for any suggestion or solution!

REPLY [3 votes]: The question can be simplified down to: does there exists naturals $a_i$ for $1\leq i \leq n$ such that
$$a_1p_1=a_2p_2=\dots=a_{n-1}p_{n-1}=a_np_n-1$$
This is in fact true. Let $L=p_1p_2\cdots p_{n-1}$. Clearly, $p_i|L$ for $1\leq i\leq n-1$ and that $\gcd\left(p_n,L\right)=1$. Thus, there exists $x,y\in\mathbb{N}$ such that
$$xp_n-yL=1\Rightarrow yL=xp_n-1$$
Now, for $1\leq i\leq n-1$, define
$$a_i=y\frac{L}{p_i}$$
and 
$$a_n=x$$
Then
$$a_ip_i=yL\text{ for }1\leq i \leq n-1$$
and
$$a_np_n=xp_n=yL+1$$
Having proved there exists naturals such that
$$a_1p_1=a_2p_2=\dots=a_{n-1}p_{n-1}=a_np_n-1$$
is true, we move on to the end of the proof. As @John_Omiela noted in his comment, if you have one solution you have infinite solutions. One solution is
$$x_i=(n-1)^{a_i}$$
Then
$$x_1^{p_1}+x_2^{p_2}+\cdots +x_{n-1}^{p_{n-1}}=(n-1)^{a_1p_1}+(n-1)^{a_2p_2}+\cdots+(n-1)^{a_{n-1}p_{n-1}}$$
$$=(n-1)(n-1)^{yL}=(n-1)^{yL+1}$$
while
$$x_n^{p_n}=(n-1)^{a_np_n}=(n-1)^{yL+1}$$
Thus, equality has been established and we are done.