Linear Independence of Unipotent Upper Triangular Matrices Let $A\ne I$ be an upper triangular unipotent matrix in $\mathbb{GL}_n(k)$, where $k$ is a field of characteristic $p$, i.e., $A$ is an upper triangular matrix all of whose diagonal elements are $1$.  Suppose further that $A^p=I$.  Are the matrices $A^i$ for $i=1,\ldots,p$ linearly independent over $k$?
This is true for $p=2$, and I suspect (and hope) it is true in general.  It seems simple enough to prove/disprove, I just can't seem to find the proper argument.  I can note the following:
$$
\sum_{i=1}^{p}c_iA^i=0\quad\Longrightarrow\quad\sum_{i=1}^{p}c_i=0
$$
for $c_i\in k$.  This implication is obtained by considering only the diagonal elements.
 A: (In general $A^p = I$ only if $n \le p$, but OP has now addressed this in his revision.)
Then, the answer in general is no. Suppose $p \ge 3$, and consider a matrix $A = I + N$, where $N^2 = 0 \ne N$. (Suppose $N$ has only one non-zero entry.)
Then $A^0 = A^p = I, A = I + N, A^2 = I + 2 N$ are linearly dependent,
$$
I - 2 (I+N) + (I + 2 N) = 0.\tag{rel}
$$
Addendum Your statement is true, however, if and only if $A = I + N$, where $N^{p-1} \ne 0 = N^p$. This is because $I, N, \dots, N^{p-1}$ are linearly independent: if
$$
a_0 I + a_1 N + \dots + a_{p-1} N^{p-1} = 0,
$$
multiply by $N^{p-1}$ to see $a_0 = 0$, then by $N^{p-2}$ to see $a_1 = 0$, etc.
It follows that $I, A, A^2, \dots , A^{p-1}$ are linearly independent, just expand the binomials. The same binomial expansion show that these elements are not linearly independent if $N^{p-1} = 0$.
Addendum$^{\textbf2}$ Perhaps I should add that you can see this from the point of view of minimal polynomials and simple extensions.
In the first example, note that since $N^2 = 0 \ne N$, the minimal polynomial of $A$ is $(x -1)^{2}$, hence (rel).
In the second example, since $N^{p-1} \ne 0 = N^p$, the minimal polynomial of $A$ is $(x-1)^{p}$, so $I, A, \dots, A^{p-1}$ are linearly independent.
When $N^{p-1} = 0$,  then $A$ is a root of $(x-1)^{p-1}$, so
$$
A^{p-1} + A^{p-2} + \dots + A + I = 0,
$$
shows that  $I, A, A^2, \dots , A^{p-1}$ are linearly dependent.
A: Everything involving $p$, including the characterestic of the field, is a red herring. For a unipotent matrix $A=I+N$ with $N$ nilpotent (never mind whether it is upper triangular) the powers $I=A^0,\ldots,A^{i-1}$ are linearly independent if and only if $i\leq d$ where the minimal polynomial of $N$ is $X^d$ (and that of $A$ is $(X-1)^d$), which means that $d$ is the size of the largest block in the Jordan normal form of $N$ (which always exists for nilpotent matrices). That's the whole story.
This means you cannot conclude without any hypothesis that forces large Jordan blocks; you only have $A\neq I$ which just ensures $d\geq2$. The condition $A^p=I$ means (given that $p$ is the characteristic) that $(A-I)^p=0$ so $d\leq p$, which doesn't help.
Since your question confusingly leaves out $A^0$ from the list but includes $A^i$, I'll need show that makes no difference. But $A^1,\ldots,A^i$ are obtained from $A^0,\ldots,A^{i-1}$ by multiplying by the invertible matrix $A$, so that does not change linear indepenence.
