Suppose the opposite. So, $abc\neq 0\implies a,b,c\in \mathbb{R*}$. We know that $a^{n+1}+b^{n+1}= c^{n+1}$, $a^{n+2}+b^{n+2}= c^{n+2}$, $a^{n+3}+b^{n+3}= c^{n+3}$ (*) By (*), we know: $a^{n+2}+b^{n+2}= c^{n+2}$ $\cdot (a+b)$ $\implies a^{n+3}+b^{n+3}+ a^{n+2}b+b^{n+2}a= c^{n+2}(a+b)$ $\implies c^{n+3}+ ab( a^{n+1}+b^{n+1})-c^{n+2}(a+b)= 0$ $\implies c^{n+3}-c^{n+2}(a+b)+abc^{n+1}=0$. Again, as $c\neq 0$ $\implies c^2-c(a+b)+ab= 0$. So, we now have an equation of degree 2 in function of $"c"$. Solving, we know $c= \frac{(a+b)\pm\sqrt{(a+b)^2-4ab}}{2}\implies c= \frac{(a+b)\pm\sqrt{(a-b)^2}}{2}$. $\therefore c= \frac{(a+b)\pm(a-b)}{2}\implies c= a$ or $c= b$. In both cases, replacing in (*), we will have one of $a,b$ or $c$ equals 0, contradiction! Therefore, our supposition is false and $abc=0$, QED.

$a^{n-1}+b^{n-1}=c^{n-1}, a^{n}+b^{n}=c^{n}, a^{n+1}+b^{n+1}=c^{n+1} $ $c^{2n}=a^{2n}+b^{2n}+2a^nb^n = (a^{n-1}+b^{n-1})(a^{n+1}+b^{n+1})=a^{2n}+b^{2n}+a^{n-1}b^{n-1}(a^2+b^2) \to a^{n-1}b^{n-1}(a-b)^2=0$ If $ab \neq 0 \to a=b$ then $c^n=2a^n,c^{n+1}=2a^{n+1} \to a=c=0$ So $abc=0$

Notice that it suffices to prove that one among ${a, b, c}$ is equal to 0. Suppose $a^n+b^n= c^n$, $a^{n+1}+b^{n+1}= c^{n+1}$ and $a^{n+2}+b^{n+2}= c^{n+2}$. We get $b^{n+1}= c^{n+1} - a^{n+1} = c^{n+1} - a(c^n - b^n)$ $\implies b^n(b-a) = c^n(c-a) (*) $ Analogously, for $a^{n+2}+b^{n+2}= c^{n+2}$ we get $b^n(b^2-a^2) = c^n(c^2-a^2)$ $\implies b^n(b+a)(b-a) = c^n(c+a)(c-a)$ If $b^n(b-a) \neq 0$, we use $(*)$ and divide in both sides to get $b+a = c+a \to b=c$ and then $a^n+b^n=b^n \implies a=0$, as desired. Else, we must have $b=0$ or $b-a=0$ and $c=0$ or $c-a=0$. Since $b=0$ or $c=0$ finishes the problem, it just remains to notice that if $b-a=c-a=0$, we get $a=b=c$ and then $a^n = 2a^n \implies a=0$, as desired.