Prove there isn't a non-zero nilpotent element of a finite ring

Question

Let $(A, +, \cdot)$ be a finite ring with $n$ elements having the property: $x^n \ne1, \forall x \ne 1, x \in A$. Prove $0$ is the only nilpotent element of $A$.

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My attempt

Suppose there is $y \in A, y \ne 0$ so that $y^k = 0$ for some $k \in \mathbb{N}-\{0, 1\}$, $k$ minimal. The idea might be to contradict the minimality of $k$ by constructing a $k' \lt k$, having the same property as $k$ has, but I don't see how to do it.

Answer 1

I've got the proof. Using Newton's binomial : $(y^{k-1} + 1)^n = y^k \cdot a + ny^{k-1} + 1 = 1 \tag 1$

Therefore $y^{k-1}=0$ contradiction with $k$ minimality.

I followed an idea provided by @Mathematician 42