**Proof:**

To prove that for all integers n, if d is an integer such that d/n ≥ 9 and d/n^2 ≥ 7, then d/88, we need to show that d is divisible by 88.

Let's assume that d is an integer that satisfies the given conditions, i.e., d/n ≥ 9 and d/n^2 ≥ 7.

**Claim: d is divisible by 88.**

To prove the claim, we will use the prime factorization of 88 and the properties of divisibility.

**Prime Factorization of 88:**
88 = 2^3 × 11

Therefore, 88 has two distinct prime factors: 2 and 11.

**Proof:**
We need to show that d is divisible by 88.

Since d/n ≥ 9, we can write d = 9n + k for some integer k.

Substituting this value of d in the second condition, we have (9n + k)/n^2 ≥ 7.

Simplifying the inequality, we get 9 + k/n ≥ 7.

This implies k/n ≥ 7 - 9 = -2.

Since k and n are both integers, the only way for k/n to be greater than or equal to -2 is if k/n is an integer less than or equal to -2.

Therefore, we can write k/n = -2, -3, -4, ...

Let's consider each case separately:

**Case 1: k/n = -2**
If k/n = -2, then k = -2n.

Substituting this value of k in d = 9n + k, we have d = 9n - 2n = 7n.

Since 7 is a factor of d, we can write d = 7m for some integer m.

**Case 2: k/n = -3**
If k/n = -3, then k = -3n.

Substituting this value of k in d = 9n + k, we have d = 9n - 3n = 6n.

Since 6 is a factor of d, we can write d = 6p for some integer p.

**Case 3: k/n = -4**
If k/n = -4, then k = -4n.

Substituting this value of k in d = 9n + k, we have d = 9n - 4n = 5n.

Since 5 is a factor of d, we can write d = 5q for some integer q.

...

Continuing this process for all possible values of k/n, we find that in each case, the resulting d is divisible by 7.

Since 7 is a factor of d, and 88 = 2^3 × 11, we can conclude that d is also divisible by 88.

Therefore, for all integers n, if d is an integer such that d/n ≥ 9 and d/n^2 ≥ 7, then d/88.