<thinking>
To prove this statement, we need to follow these steps:

1. Understand the notation:
   - T is a linear operator
   - N(T^k) is the null space of T^k
   - R(T^k) is the range of T^k

2. Recall key properties of linear operators:
   - For any linear operator T, N(T^k) ⊆ N(T^(k+1)) and R(T^(k+1)) ⊆ R(T^k)
   - The dimensions of these spaces can only increase or decrease by a finite amount

3. Consider the ascending and descending chains of subspaces:
   N(T^0) ⊆ N(T^1) ⊆ N(T^2) ⊆ ...
   ... ⊇ R(T^2) ⊇ R(T^1) ⊇ R(T^0)

4. Use the fact that the vector space is finite-dimensional:
   - This means that these chains must eventually stabilize
   - There exists a k such that N(T^k) = N(T^(k+1)) and R(T^k) = R(T^(k+1))

5. Prove that for this k, N(T^k) ∩ R(T^k) = {0}:
   - Let x ∈ N(T^k) ∩ R(T^k)
   - Since x ∈ R(T^k), there exists y such that T^k(y) = x
   - Since x ∈ N(T^k), T^k(x) = 0
   - But T^k(x) = T^k(T^k(y)) = T^(2k)(y) = 0
   - This means y ∈ N(T^(2k)) = N(T^k), so T^k(y) = 0
   - Therefore, x = T^k(y) = 0

<reflection>
This proof relies on the fundamental theorem of linear algebra, which states that for any linear operator T on a finite-dimensional vector space, there exists a k such that N(T^k) = N(T^(k+1)) and R(T^k) = R(T^(k+1)). This theorem is indeed correct and widely used in linear algebra. The proof that N(T^k) ∩ R(T^k) = {0} for this k is also logically sound, using the properties of null spaces and ranges.
</reflection>
</thinking>

<output>
The proof that there exists a nonnegative integer k such that N(T^k) ∩ R(T^k) = {0} for a linear operator T on a finite-dimensional vector space can be summarized as follows:

1. Consider the ascending chain of null spaces N(T^0) ⊆ N(T^1) ⊆ N(T^2) ⊆ ... and the descending chain of ranges ... ⊇ R(T^2) ⊇ R(T^1) ⊇ R(T^0).

2. Due to the finite dimensionality of the vector space, these chains must eventually stabilize. There exists a k such that N(T^k) = N(T^(k+1)) and R(T^k) = R(T^(k+1)).

3. For this k, we can prove that N(T^k) ∩ R(T^k) = {0}:
   Let x ∈ N(T^k) ∩ R(T^k).
   Then there exists y such that T^k(y) = x.
   Since x ∈ N(T^k), T^k(x) = 0.
   But T^k(x) = T^k(T^k(y)) = T^(2k)(y) = 0.
   This means y ∈ N(T^(2k)) = N(T^k), so T^k(y) = 0.
   Therefore, x = T^k(y) = 0.

Thus, we have proved that there exists a nonnegative integer k such that N(T^k) ∩ R(T^k) = {0}.
</output>