Induction after $\max(m,n)$. In fact we can WLOG $m>n$, and we will do induction after $m$. Suppose $m=1$. Then the result is obvious $\gcd(a_1,a_0) = a_1 = a_{\gcd(1,0)}$. Suppose that we have proved for all pairs $(m',n')$ such that $m>m'>n'$. Then $$\gcd(a_m,a_n) = \gcd(P^{(m-n)}(a_n),a_n) = \gcd(P^{(m-n)}(0),a_n)= \gcd(a_{m-n},a_n),$$ which by induction is $a_{\gcd(m-n,n)}$, which obviously equals $a_{\gcd(m,n)}$.

sorry, but who can explain for me why $gcd(P^{(m - n)}(a_n);a_n) = gcd(P^{(m - n)}(0);a_n)$ (and I think that because $a_0 = 0$ so $P^{(m - n)}(0) = 0$) is it wrong thanks

$P^{m-n}(0)$ certainly need not equal 0. The reason is that $a_n=a_n-0|P(a_n)-P(0)|P^{(2)}(a_n)-P^{(2)}(0)|\cdots|P^{(m-n)}(a_n)-P^{m-n}(0)$, so that if $d|a_n$ and $d|P^{(m-n)}(a_n)$ then $d|P^{(m-n)}(0)$, and similarly, if $d|a_n$ and $d|P^{(m-n)}(0)$ then $d|P^{(m-n)}(a_n)$.