How do I find the oxidation number of Mn in KMnO4 step by step?

Use the rule that the sum of all oxidation numbers in a neutral compound equals zero. In KMnO4, potassium (an alkali metal) is always +1, and oxygen is usually -2. Let the oxidation number of manganese be x. Now write the sum: (+1 for K) + (x for Mn) + (4 times -2 for the four oxygens) = 0. That gives 1 + x - 8 = 0, so x - 7 = 0, which means x = +7. So manganese is in the +7 oxidation state in KMnO4, its highest possible. This is why KMnO4 is such a strong oxidising agent, manganese readily drops from +7 to lower states like +2. Always fix the known elements first (K and O here), then solve for the unknown.