jmc

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 $$p(x)=c_{2005}{x}^{2005}+c_{2004}{x}^{2004}+\cdots +c_{1}x.$$If we multiply the $n$th equation by $c_n,$ then we get $$\begin{array}{ccccccccccc}{a}_1\cdot c_1\cdot 1& +& a_2\cdot c_1\cdot 2& +& a_3\cdot c_1\cdot 3& +& \cdots & +& a_{2005}\cdot c_1\cdot 2005& =& 0,\\ a_1\cdot c_2\cdot 1^2& +& a_2\cdot c_2\cdot 2^2& +& a_3\cdot c_2\cdot 3^2& +& \cdots & +& a_{2005}\cdot c_2\cdot 2005^2& =& 0,\\ a_1\cdot c_3\cdot 1^3& +& a_2\cdot c_2\cdot 2^3& +& a_3\cdot c_3\cdot 3^3& +& \cdots & +& a_{2005}\cdot c_3\cdot 2005^3& =& 0,\\ & & & & & & & & & \dots ,& \\ a_1\cdot c_{2004}\cdot 1^{2004}& +& a_2\cdot c_2\cdot 2^{2004}& +& a_3\cdot c_{2004}\cdot 3^{2004}& +& \cdots & +& a_{2005}\cdot c_{2004}\cdot 2005^{2004}& =& 0,\\ a_1\cdot c_{2005}\cdot 1^{2005}& +& a_2\cdot c_2\cdot 2^{2005}& +& a_3\cdot c_{2005}\cdot 3^{2005}& +& \cdots & +& a_{2005}\cdot c_{2005}\cdot 2005^{2005}& =& c_{2005}.\end{array}$$Note that the terms in the $k$th column add up to $p(k).$ Thus, $$a_{1}p(1)+a_{2}p(2)+a_{3}p(3)+\cdots +a_{2005}p(2005)=c_{2005}.$$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 $$p(x)=x(x-2)(x-3)\cdots (x-2005).$$Then $p(1)=2004!$ and $p(k)=0$ for $k=2,$, 3, $\dots ,$ 2005, so $$2004!\cdot a_1=1.$$Hence, $a_1=\boxed{\frac{1}{2004!}}.$