There is a unit u and primes $\pi_{j}=a_{j}+b_{j}i$ in $\mathbb{Z}[i]$ with $a_{j}>0, b_{j}>0$ and $7+i = u\pi_{1}\dots\pi_{k}$
$\mathbb{Z}[i]$ has four units: $i,-i,1,-1$. The product of the primes is also in $\mathbb{Z}[i]$, so : $7+i=u\pi… \pi_{k} = i(a+bi) = -b+ai$. This can't be the case since then it would follow that : $a=1, b=-7$ but since $b_{j}> 0$ i therefore can't be that unit.
For -i : we get -i(a+bi)= -ai+b . With this it follows that $a=-1, b= 7 $, therefore since $a_{j}>0$ this also can not be true.
For 1: 1(a+bi)= a+bi , a=7, b=1. This could be true, so now it is to show that 7+i is prime in $\mathbb{Z}[i]$. We can see directly that 7+i is irreducible over $\mathbb{Z}[i]$ and that means it must also be prime since $\mathbb{Z}[i]$ is a UFD. Therefore 1 is that unit.
I believe this can't be right, because I didn't use the indices at all. Does anybody see the right way? Please do tell.