How can I calculate $(1+i)^{5404}$? I saw a pattern while evaluating some other powers of similar complex number so I tried to calculate the above question by expanding it, please tell me if it is correct...?
$(1+i)^{2} = 2i$
$(1+i)^{4}$ = $(2i)^2$ = $-4$
$(1+i)^{8}  =  (-4)^2 = 16$
similarly,
$(1+i)^{5404}$ = $(1+i)^{4096}$ $(1+i)^{256}$ $(1+i)^{32}$ $(1+i)^{16}$ $(1+i)^{4}$
$(1+i)^{5404}$ = $(256)^{256} (256)^{16} (256)^2 (16)^2 (-4)$ 
But $(256)^{256} = \infty$
Now how can I solve it? Is my method wrong?
 A: Use Euler's Formula $$e^{i\theta}=\cos (\theta) + i\sin (\theta)$$
since $1+i=\sqrt{2} (\cos\frac{\pi}{4} +i\sin\frac{\pi}{4})= \sqrt{2} e^{i\frac{\pi}{4}}$, so $(1+i)^{5404} =(\sqrt{2} e^{i\frac{\pi}{4}})^{5404}= 2^{2702} e^{i\cdot 1351\pi}= 2^{2702} e^{i\pi}= -2^{2702}$.
A: Here is a general method without using Euler's formula, and formulas for this expansion for any $k$ exponent.
Since we can view multiplication of $i$ as a rotation by 90 degrees, which is
periodic with return to the point of origin after $4$ rotations, we see that $i$ is an element of order 4 (if you've seen a little algebra). 
This gives us:
$i^{n}=i$ whenever $n\cong 1 (\mod 4)$
$i^{n}=-1$ whenever $n\cong 2 (\mod 4)$
$i^{n}=-i$ whenever $n\cong 3 (\mod 4)$
$i^{n}=1$ whenever $n\cong 0 (\mod 4)$
Breaking up by these cases, and searching for a pattern from a
few computations you can conjecture
the following formulas:
If $k\cong 1 (\mod 4)$, we have:
    \begin{equation*}
    (1+i)^{k}=(-4)^{\lfloor{k/4}\rfloor}(1+i)
\end{equation*}
  if $k\cong 2 (\mod 4)$, we have:
  \begin{equation*}
  (1+i)^{k}=(-4)^{\lfloor{k/4}\rfloor}2i
  \end{equation*}
  if $k\cong 3 (\mod 4)$, we have:
  \begin{equation*}
  (1+i)^k=(-4)^{\lfloor{k/4}\rfloor}(2i-2)
  \end{equation*}
  if $k\cong 0 (\mod 4)$, we have:
  \begin{equation*}
  (1+i)^k=(-4)^{k/4}
  \end{equation*}
All of which can be proved by induction without too much trouble.  
A: Thanks everyone... Here is the solution as I actually wanted to calculate it. But the main problem was I used calculator instead of my pen!
$(1+i)^{4} = -4$
   &
$(1351)(4) = 5404$
$(1+i)^{5404} = (-4)^{1351}$
$(-4)^{1351} = (-4)^{1350} (-4)$
$(-4)^{1351} = (4)^{675} (4)^{675} (-4)$
$(-4)^{1351} \approx (-2.2) (10)^{812}$
$(1+i)^{5404} \approx (-2.2) (10)^{812}$
