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Let $A$ be an $ntimes n$ complex matrix, and suppose that $A^nneq 0$. Prove that $A^kneq0;forall;kinmathbbN$.

$rule18cm1pt$

My arguments $textbf1$:

Let $O_ntimes n$ is the zero matrix.

Since $A^nneq O_ntimes nimplies det(A^n)neq0implies (det(A))^nneq0 oversetdet(A)inmathbbCimplies det(A)neq0$. $Aneq O_ntimes n$ as $det(O_ntimes n)=0.$ Suppose $A^k=O_ntimes n$ for some $kinmathbbN$. $kneq n$ else this contradicts the statement.

$k<n$ then, $$A^n=A^kcdot A^n-k=O_ntimes n;text as; A^k=O_ntimes n RightarrowLeftarrow A^nneq O_ntimes n$$

$k>n$ then, $$beginalignA^k&=A^ncdot A^k-n\ &= A^ncdot A^ncdot A^k-2n tagif $k-n>n$\ & quadqquadvdots\
&=A^ncdot A^ncdots A^ncdot A^k-jntag1endalign$$
such that $k-jnleq n$.

If $k-jn=n$ then $(1)implies$ $A^k=(A^n)^m$ for some $minmathbbN$.
$$0=det(A^k)=det((A^n)^m)=(det(A^n))^mimplies det(A^n)=0RightarrowLeftarrow det(A^n)neq0$$

If $k-jn<n$ then, $(1)implies$ $A^k=(A^n)^mcdot A^k-jn$ for some $minmathbbN$.

$$displaystyle 0=det(A^k)=det((A^n)^mcdot A^k-jn)=(det(A^n))^mcdot (det(A))^k-jn$$ $$oversetdet(A^n)neq0implies (det(A))^k-jn=0implies det(A)=0 RightarrowLeftarrow det(A)neq0$$

Hence the statement.
$rule18cm0.5pt$
My arguments $textbf2$:

Since $A^nneq O_ntimes nimplies det(A^n)neq0implies (det(A))^nneq0 oversetdet(A)inmathbbCimplies det(A)neq0$. $Aneq O_ntimes n$ as $det(O_ntimes n)=0.$ Suppose $A^k=O_ntimes n$ for some $kinmathbbN$. $kneq n$ else this contradicts the statement.

$$det(A^k)=0oversetdet(A)inmathbbCimplies det(A)=0RightarrowLeftarrow det(A)neq0$$

$rule18cm1pt$

I feel that my arguments in $textbf1$ are circular while in $textbf2$ are insufficient. Could anyone tell me which parts can be ommitted in $textbf1$ and comment about $textbf2$?

In $mathbb C$ all matrices are triangularizable. Therefore, there is a basis s.t. $A=D+N$ where $D$ is diagonal and $N$ nilpotent. If $A^nneq 0$, then $Dneq 0$, and thus $A^kneq 0$ for all $k$.

$Aneq 0$ do not implies that $det(A)neq 0$. For example, $beginpmatrix1&0\0&0endpmatrix$ and many other matrices...

By contradiction assume that for some $k$ we have $A^k=0$ so the polynomial $x^k$ annihilates $A$ and so the minimal polynomial is $x^p$ for some $p$ and the characteristic polynomial is $x^n$. Using Cayley-Hamilton theorem we get $A^n=0$ which is a contradiction.

Both arguments are wrong at the very start: $A^nne0$ does not imply $det(A^n)ne0$. (Counterexample: $A=beginbmatrix1&0\0&0endbmatrix$.)

The correct arguments I see all use somewhat more advanced concepts.

Hint: Suppose that $A^k=0$. Show that the minimal polynomial for $A$ must be $m(t)=t^j$ for some $j$. Show that $jle n$; hence $A^j=0$ implies $A^n=0$.