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to the contrary, there is an integer that has two distinct prime factorizations. Let be the least such integer and write , where each and is prime. We see that divides , so divides some by Euclid's lemma. Without loss of generality, say divides . Since and are both prime, it follows that . Returning to our factorizations of , we may cancel these two factors to conclude that . We now have two distinct prime factorizations of some integer strictly smaller than , which contradicts the minimality

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