Python 3, math solution that follows hints
Hints followed\n\n\nclass Solution:\n    def abbreviateProduct(self, left: int, right: int) -> str:\n        # hints followed\n        left = max(2, left)\n    
Hints followed\n\n```\nclass Solution:\n    def abbreviateProduct(self, left: int, right: int) -> str:\n        # hints followed\n        left = max(2, left)\n        mod = 10 ** 5\n        p = 1 # 10**5 modulo of the product to represent the lowest 5 digits\n        lg = 0 # cumulates log10 of each number left to right, for the highest 5 digits\n        c2 = 0 # occurrences of prime factor 2\n        c5 = 0 # occurrences of prime factor 5\n        \n        for i in range(left, right + 1):\n            lg += log10(i)\n            while i > 1 and not (i % 2):\n                c2 += 1\n                i //= 2\n            while i > 1 and not (i % 5):\n                c5 += 1\n                i //= 5\n            p *= i\n            if p > mod:\n                p %= mod\n        \n        e = min(c2, c5)\n        es = \'e\' + str(e)\n        c2 -= e\n        c5 -= e\n        if not c2:\n            p *= pow(5, c5, mod)\n        else:\n            p *= pow(2, c2, mod)\n        if p > mod:\n            p %= mod\n        ps = str(p)\n        \n        lg = lg - e\n        if lg <= 5:\n            return ps + es\n        \n        # more than 5 digits to display, fill ps with leading 0s as need\n        if len(ps) < 5:\n            ps = \'0\' * (5 - len(ps)) + ps\n        \n        # digits to display <= 10\n        lg -= 5\n        if lg < 5:\n            return str(10**lg)[:ceil(lg)] + ps + es\n        \n        # digits to display > 10, abbreviate the middle\n        lg -= (ceil(lg) - 5)\n        return str(10**lg)[:5] + \'...\' + ps + es\n```
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