TITLE: Is it true? $(1+ta_1)(1+ta_2)\cdots(1+ta_n)=1, \;\forall t\in{\Bbb R}$ ⟺ $a_1=a_2=...=a_n=0$
QUESTION [5 upvotes]: $(1+ta_1)(1+ta_2)...(1+ta_n)=1$ $\forall t\in R$ if and only if $a_1=a_2=...=a_n=0$
Is it true? If yes, how to prove it?
I already gave a proof when $n=3$.
$1=(1+ta_1)(1+ta_2)(1+ta_3)=1+(a_1+a_2+a_3)t+(a_1a_2+a_1a_3+a_2a_3)t^2+(a_1a_2a_3)t^3$
It is equivalent to
$a_1+a_2+a_3=0$
$a_1a_2+a_1a_3+a_2a_3=0$
$a_1a_2a_3=0$
And this gives
$a_1=a_2=a_3=0$
Is it possible to generalize it for arbitrary $n$?
$(1+ta_1)(1+ta_2)\cdots(1+ta_n)=1$, $\forall{t}$ $\in$ $\Bbb{R}$ ⟺ $a_1=a_2=\cdots=a_n=0$

REPLY [2 votes]: I've just found a new way to prove this.
It's not hard to see that the coefficients of polynomial are all zeros. When $n=4$, that is to say,
$a_1+a_2+a_3+a_4=0$
$a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4=0$
$a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4=0$
$a_1a_2a_3a_4=0$
By the last equality, at least one $a_i(i=1,2,3,4)$ is zero. Let's assume $a_1=0$.
Then by the second last equality, $a_2a_3a_4=0$. Let's assume $a_2=0$.
Then by the third last equality, $a_3a_4=0$. Let's assume $a_3=0$.
Then by the first equality, $a_4=0$.