Question 897746
<pre>
{{{i^2014!+i^2013!+i^2012!+i^2011!+i^2010!+""*""*""*""+i^3!+i^2!+i^1!}}}

All the factorials in the exponents are multiples of 4 except the last three.

When i is raised to the power of any multiple of 4, the result is 1. 

That's because {{{i^(4n)=i^(2*2n)=(i^2)^(2n)=(-1)^(2n)=((-1)^2)^n=1^n=1}}} 
 
So the above becomes:

{{{1+1+1+1+1+""*""*""*""+1+i^3!+i^2!+i^1!}}}

There are 2014-3 or 2011 1's, and for the last three terms

{{{i^3!=i^(3*2*1)=(i^2)^3=i^2*i=(-1)i=-i}}}
{{{i^2!=i^(2*1)=i^2=-1}}}
{{{i^1!=i^1=i}}}

So the sum is {{{2011+(-i)+(-1)+i=2010}}}

Edwin</pre>