Finding the last two digits of the given number. Question:

Given $N=2^5+2^{{5}^{2}}+2^{{5}^{3}}+2^{{5}^{4}}... 2^{{5}^{2015}}$
  Written in the usual decimal form, find the last two digits of the number $N$.  

My attempt: 
We know that every exponential number repeat its last,last two,last third , $\ldots$ digits. For eg.: unit digits of $3^n$ repeats itself as $3,9,7,1,3,9,7,1...$ On applying this for last two digits of $2^n$ we get
$02,04,08,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44,88,76,52$ and $04,08...$ again.  
This repetitive series has $21$ numbers with first one(i.e.$1$) occurring only once.
Also all $n$ in $2^n$; $5,125,625...$ when divided by $20$ give $5$ as remainder so the last two digits will be $32,16,16,16...16$ with $16$ occurring $2014$ times which gives $32+32224=32256$ which means last two digits will be $56$ but when doing the same for last digit, I get $0$ which is a contradiction.  
I can't find what mistake I did.
Thanks for finding my mistake.
 A: We need to evaluate
$$2^5+2^{{5}^{2}}+2^{{5}^{3}}+2^{{5}^{4}}... 2^{{5}^{2015}} \mod {100}$$
and


*

*$2^5=32 \mod {100}$

*$2^{5^2}=2^{10}2^{10}2^{5}\equiv24\cdot 24\cdot 32\equiv32 \mod {100}$

*$2^{5^3}=(2^{10}2^{10}2^{5})^5\equiv 32^5 \equiv 32\mod {100}$

*...


therefore
$$2^5+2^{{5}^{2}}+2^{{5}^{3}}+2^{{5}^{4}}... 2^{{5}^{2015}} \equiv2015\cdot 32\equiv 80\mod {100}$$
A: Note that:
$$02,\underbrace{04,08,16,32,64,\cdots,76,52}_{20}, \underbrace{04,08,16,32,64,\cdots,76,52}_{20},\cdots$$
To find the last two digits of $n=5,25,125,...$, you must consider: $n-1$ mod $20$:
$$\begin{align}5-1\equiv4 \pmod{20} \Rightarrow 32\\
25-1\equiv4 \pmod{20} \Rightarrow 32\\
125-1\equiv4 \pmod{20} \Rightarrow 32\end{align}$$
Alternatively, you can consider the last two digits of $(2^5)^m=32^m:$
$$\color{red}{32};24;68;76;\color{red}{32};24;68;76;\color{red}{32};24;68;76;...$$
Since $5m=5^n \Rightarrow m=5^{n-1}=1;5;25;125;... \Rightarrow m\equiv 1 \pmod 4$, then:
$$\underbrace{32+32+\cdots+32}_{2015}=32\cdot 2015=64480\equiv 80 \pmod{100}.$$
