jmc

Prime factorize 715 to find $715=5\cdot 11\cdot 13$. The only ways to write 715 as a product of positive integers greater than 1 are the distinct ways of grouping the prime factors: $$\begin{array}{rl}(5)\cdot (11)\cdot (13)& =5\cdot 11\cdot 13\\ (5\cdot 11)\cdot 13& =55\cdot 13\\ 5\cdot (11\cdot 13)& =5\cdot 143\\ (5\cdot 13)\cdot 11& =65\cdot 11\text{, and}\\ (5\cdot 11\cdot 13)& =715,\end{array}$$ where the last one is a product with only one factor. Since the letters cannot represent numbers greater than 26, only $5\cdot 11\cdot 13$ 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 $\boxed{\text{MAKE}}$.