Prove that $\alpha$ is a cyclic vector for $N$. Let V be a $n$-dimensional vector space over the field $F$, and let $N$ be a nilpotent linear operator on $V$. Suppose $N^{n-1}\neq0$, and let $\alpha$ be a non-zero vector in $V$ s.t $N^{n-1}\alpha\neq0$. Prove that $\alpha$ is a cyclic vector for $N$.
What I have done is taking on a contrary  $\alpha$ is not a cyclic vector. $\exists k<n-1$ s.t $T^k \alpha=c_0\alpha+c_1T\alpha+\cdots+T^{k-1}\alpha$. Then applying $T$ on it again & again & putting  $T^k \alpha=c_0\alpha+c_1T\alpha+\cdots+T^{k-1}\alpha$. I manage to show that $c_i=0$ $\forall i=1(1)k-1$. But this method is cumbersome. So can anyone give me any other short proof?? 
 A: One "shorter" proof I can think of is very similar to what you did, but directly proving that $\;\{\alpha\,,\,T\alpha\,,\ldots, T^{n-1}\alpha\}\;$ is linearly independent: so suppose $\;a_0,...,a_{n-1}\in F\;$ are such that
$$\sum_{k=0}^{n-1}a_kT^k\alpha=0\implies 0=T^{n-1}(0)=T^{n-1}\left(\sum_{k=0}^{n-1}a_kT^{n+k-1}\alpha\right)=a_0T^{n-1}\alpha$$
since $\;T^{n+k-1}=0\;,\;\;\forall\,k\ge 1\;$ , and since $\;T^{n-1}\alpha\neq 0\;$ we get that $\;a_0=0\;$
Go back with the same trick as above for what is left:
$$\sum_{k=1}^{n-1}a_kT^k\alpha$$
and deduce $\;a_1=0\;$ and etc. ... or you can do as follows (reductio ad absurdum): suppose the very first relation above (third line) is a minimal one, meaning: a trivial linear combination with all its coefficients different from zero, say
$$\sum_{k=i_1}^{i_r} a_{i_1}T^{i_1}\alpha=0\;,\;\;0\le i_1<i_r\le n-1\;,\;\;a_j\neq 0\;\;\forall\,\, i_1\le j\le i_r$$
and now apply $\;T^{n-i_r\;}\;$ to the above minimal lin. comb. and get that $\;a_{i_1}=0\;$ , which gives a direct and final contradiction and we're done. Much shorter, indeed.
