Write 2014^2014 in scientific notation.

2014^2014

( $2*10^3$ + $1*10^1$ + $4*10^0$ )^2014

[(10^3)(2+.01+.004)]^2014

(2.014^2014)(10^6042)

Now the challenge is to write 2.014^2014 as base 10.

2.014=10^[log(2.014)]

2.014^2014=10^[ $log(2.014)*2014$ ]

Now we want this to equal a number times ten to the x power. That # has to be between 0 and 10.

10^[log(2.014)*2014]=#(10^x)

10^[log(2.014)*2014-x]=#

When x=612, #=2.376,

So plugging it back into the expression, the expression simplifies to

(2.376*10^612)(10^6042)

2.376E6654.

By Mathematica,

$2014^{2014}=2.3755543019126293767818152900915899082687738760930226775925776222508418285734707410383776451948654680315185844528203348213179234685234841077263988192052090847017293165688937558699711617267173581471688977427274559160978642465507998293752316442829450969314926068184685720193475525391961203707338317285658761073115708796150106313365775426017170209232653108379015662325454602950249246609616766021777463815636705963690396329255376084155291336731847871295599757311739523611454623366204272656524660753594088862407763610683159883547873360375303374622450632271100870095214906790108778896134844528916720166244313375184860185140011760032645160997528351125245220534909180080282983310982141487600480676091308067987277118294628500518878351413505106343175464312259262866152351153627319236666972268159516484943011191212244227086047571201371046931305882369935600893334352471149019601177647686163657596540739150766196241932580109561710693864114882342003324941654963230770586085242088789891411299315223978551400684868445982715170212952334269422370986624950292715992552642×10^{6654}$ .