Product polynomial in $\mathbb{F}_7$ I need to compute the product polynomial $$(x^3+3x^2+3x+1)(x^4+4x^3+6x^2+4x+1)$$
when the coefficients are regarded as elements of the field $\mathbb{F}_7$.
I just want someone to explain to me what does it mean when a cofficient (let us take 3 for example) is in $\mathbb{F}_7$ ? I know that $\mathbb{F}_7= \mathbb{Z}/7\mathbb{Z}$

As others have mentioned, an integer $n$ in $\mathbb{Z}_p\equiv{n\mod p}$.  So coefficients can only be in the set $\{0,1,2,...,p-2, p-1\}$ Thus, you can knock out terms with coefficients of the form $pn$.  In your case multiples of $7\equiv{0\mod7}$.
I will add a little more of a hint to cameron's
$(a+b)^p={p\choose0}a^p+{p\choose 1}a^{p-1}b+{p\choose 2}a^{p-2}b^2+...+{p\choose p-1}ab^{p-1}+{p\choose p}b^p$
$=\frac{p!}{p!0!}a^p+\frac{p!}{(p-1)!1!}a^{p-1}b+\frac{p!}{(p-2)!2!}a^{p-2}b^2+...+\frac{p!}{1!(p-1)!}ab^{p-1}+\frac{p!}{0!p!}b^p$
$=1a^p+pa^{p-1}b+\frac{p(p-1)}{2}a^{p-2}b^2+...+pab^{p-1}+1b^p$
$=a^p+0a^{p-1}b+0a^{p-2}b^2+...+0ab^{p-1}+b^p$
$=a^p+b^p$
