1. **State the problem:** Prove by mathematical induction that $$\sum_{i=1}^n i^4 = \frac{n(n+1)(6n^3 + 9n^2 + n - 1)}{30}.$$\n\n2. **Base case (n=1):** Calculate the left side: $$\sum_{i=1}^1 i^4 = 1^4 = 1.$$ Calculate the right side: $$\frac{1 \cdot 2 \cdot (6 \cdot 1^3 + 9 \cdot 1^2 + 1 - 1)}{30} = \frac{1 \cdot 2 \cdot (6 + 9 + 1 - 1)}{30} = \frac{2 \cdot 15}{30} = 1.$$ Both sides equal 1, so the base case holds.\n\n3. **Inductive hypothesis:** Assume the formula holds for some integer $k \geq 1$, that is, $$\sum_{i=1}^k i^4 = \frac{k(k+1)(6k^3 + 9k^2 + k - 1)}{30}.$$\n\n4. **Inductive step:** Prove it holds for $k+1$. Consider $$\sum_{i=1}^{k+1} i^4 = \left(\sum_{i=1}^k i^4\right) + (k+1)^4.$$ Substitute the inductive hypothesis: $$= \frac{k(k+1)(6k^3 + 9k^2 + k - 1)}{30} + (k+1)^4.$$\n\n5. **Simplify the right side:** Factor out $(k+1)$: $$= \frac{(k+1)}{30} \left[k(6k^3 + 9k^2 + k - 1) + 30(k+1)^3 \right].$$\n\n6. **Expand inside the bracket:** First, expand $k(6k^3 + 9k^2 + k - 1) = 6k^4 + 9k^3 + k^2 - k.$ Next, expand $(k+1)^3 = k^3 + 3k^2 + 3k + 1,$ so $$30(k+1)^3 = 30k^3 + 90k^2 + 90k + 30.$$\n\n7. **Sum the terms inside the bracket:** $$6k^4 + 9k^3 + k^2 - k + 30k^3 + 90k^2 + 90k + 30 = 6k^4 + (9k^3 + 30k^3) + (k^2 + 90k^2) + (-k + 90k) + 30 = 6k^4 + 39k^3 + 91k^2 + 89k + 30.$$\n\n8. **Rewrite the sum:** $$\sum_{i=1}^{k+1} i^4 = \frac{(k+1)}{30} \left(6k^4 + 39k^3 + 91k^2 + 89k + 30\right).$$\n\n9. **Check if this matches the formula for $n = k+1$:** The formula is $$\frac{(k+1)(k+2)(6(k+1)^3 + 9(k+1)^2 + (k+1) - 1)}{30}.$$\n\n10. **Expand the polynomial inside:**\n- $(k+1)^3 = k^3 + 3k^2 + 3k + 1$\n- $(k+1)^2 = k^2 + 2k + 1$\n\nSo, $$6(k+1)^3 = 6k^3 + 18k^2 + 18k + 6,$$ $$9(k+1)^2 = 9k^2 + 18k + 9,$$ and $$ (k+1) - 1 = k.$$\n\nSum these: $$6k^3 + 18k^2 + 18k + 6 + 9k^2 + 18k + 9 + k = 6k^3 + (18k^2 + 9k^2) + (18k + 18k + k) + (6 + 9) = 6k^3 + 27k^2 + 37k + 15.$$\n\n11. **Multiply by $(k+2)$:** $$ (k+2)(6k^3 + 27k^2 + 37k + 15) = 6k^4 + 27k^3 + 37k^2 + 15k + 12k^3 + 54k^2 + 74k + 30 = 6k^4 + (27k^3 + 12k^3) + (37k^2 + 54k^2) + (15k + 74k) + 30 = 6k^4 + 39k^3 + 91k^2 + 89k + 30.$$\n\n12. **Final expression:** $$\sum_{i=1}^{k+1} i^4 = \frac{(k+1)(k+2)(6k^3 + 27k^2 + 37k + 15)}{30} = \frac{(k+1)}{30} (6k^4 + 39k^3 + 91k^2 + 89k + 30),$$ which matches the expression found in step 8.\n\n13. **Conclusion:** Since the formula holds for $n=1$ and assuming it holds for $n=k$ implies it holds for $n=k+1$, by mathematical induction the formula is true for all positive integers $n$.