Note that the $ n $th equation contains $ n $th powers, specifically $ 1^n, $ $ 2^n, $ $ \dots, $ $ 2005^n $. This makes us think of evaluating some polynomial $ p(x) $ at $ x = 1, $ 2, $ \dots, $ 2015. The question is which polynomial. So, let \[p(x) = c_{2005} x^{2005} + c_{2004} x^{2004} + \dots + c_1 x.\]If we multiply the $ n $th equation by $ c_n, $ then we get \[ \begin{array}{ccccccccccc} a_1 \cdot c_1 \cdot 1 & + & a_2 \cdot c_1 \cdot 2 & + & a_3 \cdot c_1 \cdot 3 & + & \dotsb & + & a_{2005} \cdot c_1 \cdot 2005 & = & 0, \\ a_1 \cdot c_2 \cdot 1^2 & + & a_2 \cdot c_2 \cdot 2^2 & + & a_3 \cdot c_2 \cdot 3^2 & + & \dotsb & + & a_{2005} \cdot c_2 \cdot 2005^2 & = & 0, \\ a_1 \cdot c_3 \cdot 1^3 & + & a_2 \cdot c_2 \cdot 2^3 & + & a_3 \cdot c_3 \cdot 3^3 & + & \dotsb & + & a_{2005} \cdot c_3 \cdot 2005^3 & = & 0, \\ & & & & & & & & & \dots, & \\ a_1 \cdot c_{2004} \cdot 1^{2004} & + & a_2 \cdot c_2 \cdot 2^{2004} & + & a_3 \cdot c_{2004} \cdot 3^{2004} & + & \dotsb & + & a_{2005} \cdot c_{2004} \cdot 2005^{2004} & = & 0, \\ a_1 \cdot c_{2005} \cdot 1^{2005} & + & a_2 \cdot c_2 \cdot 2^{2005} & + & a_3 \cdot c_{2005} \cdot 3^{2005} & + & \dotsb & + & a_{2005} \cdot c_{2005} \cdot 2005^{2005} & = & c_{2005}.\end{array} \]Note that the terms in the $ k $th column add up to $ p(k) $. Thus, \[a_1 p(1) + a_2 p(2) + a_3 p(3) + \dots + a_{2005} p(2005) = c_{2005}.\]Note that this holds for any constants $ c_1, $ $ c_2, $ $ \dots, $ $ c_{2005} $ we choose. Since we want $ a_1, $ we choose the coefficients $ c_i $ so that all of the terms in the equation above disappear, except for $ a_1 p(1) $. We can achieve this by setting \[p(x) = x(x - 2)(x - 3) \dotsm (x - 2005).\]Then $ p(1) = 2004! $ and $ p(k) = 0 $ for $ k = 2, $, 3, $ \dots, $ 2005, so \[2004! \cdot a_1 = 1.\]Hence, $ a_1 = \boxed{\frac{1}{2004!}} $.