Let $T : V → V$ be a linear map, where $\dim(V ) = n$, and suppose that $T^n = 0$ and that there exists a vector $v ∈ V$ with $T^{n−1}(v) \neq 0$.
Prove that the vectors $v, T(v), T^2(v), . . . , T^{n−1}(v)$ are linearly independent and that the nullity of $T$ is $1$.
Suppose to the contrary that $v, T(v), T^2(v), \dots, T^{n-1}(v)$ are linearly dependent. Then there exist coefficients $a_i$ such that $$ b_0v + b_1T(v) + \cdots + b_{n-1} T^{n-1}(v) = 0 $$ Let $i_0$ denote the lowest $i$ for which $b_i \neq 0$. Then we have $$ T^{i_0}v = \sum_{i = i_0 + 1}^n a_iT^iv $$ for some coefficients $a_i$. It follows that $$ T^{n-1} v = T^{n-1-i_0}T^{i_0}v = T^{n - 1 - i_0}\sum_{i=i_0+1}^n a_i T^iv = \\ \sum_{i=i_0+1}^n a_i T^{n + (i-(i_0+1))}v = \sum_{j=0}^{k-1} a_{j+1+i_0} \underbrace{T^{n + j}}_{=0}v = 0 $$ Which contradicts how we defined $v$.
For the second part, note that $\{T(v),T^2(v),\dots,T^{n-1}(v)\}$ are a linearly independent subset of the image of $T$.