jmc

Each group of 4 consecutive powers of $i$ adds to 0: $i+i^2+i^3+i^4=i-1-i+1=0$, $i^5+i^6+i^7+i^8=i^4(i+i^2+i^3+i^4)=1(0)=0$, and so on. Because 600 is divisible by 4, we know that if we start grouping the powers of $i$ as suggested by our first two groups above, we won't have any `extra' powers of $i$ beyond $i^{600}$. We will, however, have the extra 1 before the $i$, so: $$i^{600}+i^{599}+\cdots +i+1=(0)+(0)+\cdots +(0)+1=\boxed{1}.$$