Till 2015

$1000=2^3\cdot 5^3$ . We're solving $2015^{\ldots}\pmod{\! 1000}$ .

$2015^{\ldots}\equiv 0\pmod{\! 5^3}$ and $2015^{2014^{\ldots}}\equiv (-1)^{2014^{\ldots}}\equiv 1 \pmod{\! 2^3}$ .

Now you just use Chinese remainder theorem or other things.

1) $2015^{\ldots}\equiv 625\equiv 0\pmod{\! 5^3}$ ; $2015^{\ldots}\equiv 625\equiv 1\pmod{\! 2^3}$ ,

so $2015^{\ldots}\equiv 625\pmod{\! 1000}$ , because

$2^3,5^3\mid 2015^{\ldots}-625\iff 2^35^3\mid 2015^{\ldots}-625$ (since $(2^3,5^3)=1$ ).

2) Let $2015^{\ldots}=125k$ . Then

125k\equiv 1\iff -3k\equiv 9\stackrel{:(-3)}\iff k\equiv -3\equiv 5\pmod{\! 8} .

$\iff k=8m+5$ and $2015^{\ldots}=125(8m+5)=1000m+625$ .

In first place, I'm going to prove that $\forall n\in \mathbb{N}\setminus \{1\}, 4|2014^{n}$ .

Let $X= \{ n\in \mathbb{N}\setminus \{1\}: 4|2014^{n} \}$ . We have that $2014^2=4046196$ and $\frac{4046196}{4}=1014049$ , therefore, $4|2014^{2}$ and $2\in X$ .

Let's suppose, of induction hypothesis, that $n \in X$ . In these conditions we have that, $4|2014^{n}$ , therefore, $\exists k\in \mathbb{N}: 2014^{n}=4k$ . Let $k \in \mathbb{N}$ be in that conditions. We have that $2014^{n+1}=2014^{n}\cdot 2014= 4k\cdot 2014=4\cdot(2014k)$ , which means that $4| 2014^{n+1}$ , and so, $(n+1)\in X$ .

So $X=\mathbb{N}\setminus \{1\}$ , as wanted.

Let's prove now that $\forall n\in 4\mathbb{N}$ , the three last digits of $2015^n$ are 625.

Let $Y= \{ n\in 4\mathbb{N}:\textrm{the three last digits of} 2015^n \textrm{are} 625 \}$ .

We have that $2015^4=16485427050625$ , therefore, $4\in Y$ .

Let's suppose, of induction hypothesis, that $n \in Y$ . These means that $\exists v\in \mathbb{N}: n=4v$ . Let $v\in \mathbb{N}$ be in that conditions.

We have that $2015^{4(v+1)}=2015^{4v} \cdot 2015^4=2015^n \cdot 2015^4$ . These means that the last three digits of $2015^{4(v+1)}$ are the last three digits of $625^2$ . but we know that $625^2=390625$ , therefore, the last three digits of $2015^{4(v+1)}$ are 625, and so $4(v+1) \in Y$ .

So $Y= 4 \mathbb{N}$ .

Knowing these, we can conclude that $4|2014^{2013^{{ ...}^1} }$ . So, the last three digits of $2015^{2014^{2013^{{ ...}^1} }}$ are 625, as wanted.

I solved it as follows.. It is obvious that the last 3 digit of a no., let's say 2015, raise to a large power is only dependent on 5 raise to that large power. And by observing that
5^(odd power >= 3 )= 125 (mod.1000) and 5^(even power >= 4 )=625(mod. 1000). And since power of 5 in the question is an even no. obviously hence the answer is 625.
I hope it helped. Let me know if you have any doubts. :)