Consider $T$ = $9 \times 99 \times 999 \times 9999 \times \cdots \times \underbrace{999....9}_{2015 \:nines}$ Consider $T$ = $9  \times 99 \times 999 \times 9999 \times 
 \cdots \times \underbrace{999....9}_{2015 
\:nines}$ 

Find the last 3 digits of $T$

Advise: I wrote it down wrong the first time, it should be a product of "2015" numbers, i apologize about that, i realized my fault when i was travelling and i couldn't repare it in my cellphone.

My try

I know the last digit, i found it easily, but the struggle is with the others. I tried this:

$9  \times 99 \times 999 \times \cdots \times \underbrace{999 \ldots 9}_{2015 
\:nines}$  $= 9 \times 9(11) \times 9(111) \times \cdots \times 9(\underbrace{111 \ldots 1}_{2015 
\:ones})$

So $T$ = $9(1+11+111+ \cdots +\underbrace{111 \ldots 1}_{2015 
\:ones})$

But from here i found nothing, any hints?
Hint:

$$T\equiv9\cdot99\cdot999^{2015-2}\pmod{1000}$$

Now $999\equiv-1\pmod{1000}\implies999^{2013}\equiv(-1)^{2013}\equiv-1$

$$\implies T\equiv(10-1)(100-1)(-1)\equiv-1+10+100$$
Well, following from where you ended up:

$T = 9(1 \times 11 \times 111 \times ...)$

Notice that except for the first and second terms, the other 2013 terms are congruent to 111 (mod 1000).

So $T \equiv 9(1 \times 11\times 111^{2013}) \pmod {1000}$

We will use the property that $\varphi (n^m) = n^{m-1} \varphi (n)$

So $\varphi (10^3) = 10^2 \varphi (10) = 100 \times 4 = 400 $

Since $2013 = 400 \times 5 + 13$,

$9(1 \times 11\times 111^{2013} \equiv 9(11 \times 111^{13}) \equiv 9(11 \times 111^{10} \times 111^3) \equiv 9(11 \times (111^2 )^5 \times 1367631) \equiv 9(11 \times 12321^5 \times 631) \equiv 9(11 \times 321^5 \times 631) \equiv 9(11 \times 3408200705601 \times 631) \equiv 9(11 \times 601 \times 631) \equiv 9(4171541) \equiv 9(541) \equiv 4869 \equiv 869 \pmod {1000}$