Prime factorize 715 to find $ 715=5\cdot11\cdot13 $. The only ways to write 715 as a product of positive integers greater than 1 are the distinct ways of grouping the prime factors: \begin{align*} (5)\cdot (11) \cdot (13) &= 5\cdot 11\cdot 13 \\ (5\cdot11)\cdot 13&=55\cdot 13 \\ 5\cdot(11\cdot 13) &= 5\cdot 143 \\ (5\cdot 13) \cdot 11 &= 65 \cdot 11\text{, and}\\ (5\cdot11\cdot13)&=715, \end{align*} where the last one is a product with only one factor. Since the letters cannot represent numbers greater than 26, only $ 5\cdot11\cdot 13 $ could come from calculating the product value of a word. The 5th, 11th, and 13th letters of the alphabet are E, K, and M. Since E, K, and M do not form a word, we introduce the letter A (which doesn't affect the product since its value is 1) to form the word $ \boxed{\text{MAKE}} $.