Можете решить?(1 - 1/4)(1 - 1/9)(1 - 1/16)...(1 - 1/225)

$(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})(1-\frac{1}{25})...(1-\frac{1}{225})\\\\ \frac{2^2-1}{2^2 } * \frac{3^2-1}{3^2} * \frac{4^2-1}{4^2} * \frac{5^2-1}{5^2}*...*\frac{15^2-1}{15^2} \\\\ \frac{(2-1)(2+1)}{4}* \frac{(3-1)(3+1)}{9}*\frac{(4-1)(4+1)}{16}*...*\frac{(15-1)(15+1)}{225} \\\\ \frac{(1*2*3*4*5*...*14)*(3*4*5*6*...*16)}{4*9*16*...*225} = \frac{8}{15}$