To prove that $3^n > n^4$ for all integers $n > 7$ using mathematical induction, we follow these steps:

Step 1: Base Case

First, we verify the statement for $n = 8$:

$3^8 = 6561$ $8^4 = 4096$

Clearly, $6561 > 4096$

So, the base case holds true for $n = 8$.

Step 2: Inductive Hypothesis

Assume that for some integer $k \geq 8$, the statement $3^k > k^4$ holds true. That is, $3^k > k^4$

Step 3: Inductive Step

We need to show that $3^{k+1} > (k+1)^4$.

Starting with the left-hand side: $3^{k+1} = 3 \cdot 3^k$

Using the inductive hypothesis $3^k > k^4$: $3^{k+1} = 3 \cdot 3^k > 3 \cdot k^4$

Now, we need to show that $3 \cdot k^4 \geq (k+1)^4$ for $k \geq 8$:

$3 \cdot k^4 \geq (k+1)^4$

Expanding $(k+1)^4$: $(k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$

We need to show: $3k^4 \geq k^4 + 4k^3 + 6k^2 + 4k + 1$

Subtract $k^4$ from both sides: $2k^4 \geq 4k^3 + 6k^2 + 4k + 1$

We need to verify if this inequality holds for $k \geq 8$. For $k = 8$: $2 \cdot 8^4 = 2 \cdot 4096 = 8192$ $4 \cdot 8^3 = 4 \cdot 512 = 2048$ $6 \cdot 8^2 = 6 \cdot 64 = 384$ $4 \cdot 8 = 32$

Thus, $4k^3 + 6k^2 + 4k + 1 \leq 2048 + 384 + 32 + 1 = 2465$

Clearly, $8192 > 2465$

Therefore, $2k^4 \geq 4k^3 + 6k^2 + 4k + 1$ holds for $k = 8$. Since $k^4$, $4k^3$, $6k^2$, $4k$, and the constant 1 all grow polynomially and 2$k^4$ grows faster than the polynomial sum as $k$ increases, the inequality $2k^4 \geq 4k^3 + 6k^2 + 4k + 1$ will hold for all $k \geq 8$.

Thus, by induction, we have proven that $3^n > n^4$ for all integers $n > 7$.

Would you like more details or have any questions? Here are some more questions you might consider:

1. How do you prove the base case in an induction proof?
2. What are some common pitfalls in mathematical induction?
3. How does the inductive hypothesis work in a proof?
4. What is the importance of the inductive step?
5. Can induction be used to prove inequalities in general?

Tip: Always carefully check your base case and ensure your inductive step correctly applies the inductive hypothesis.