= Solution
For $n\geq1$, let $K_n=\ker(\phi^n)$, where the power denotes repeated composition of the <ring endomorphism>. These <ideals> form an ascending chain:
$$
K_1\subseteq K_2\subseteq K_3\subseteq\cdots.
$$
The <ascending chain condition> for a <Noetherian ring> gives $K_N=K_{N+1}$ for some $N\geq1$. Take $a\in\ker\phi$. Since $\phi$ is <surjective>, its iterate $\phi^N$ is <surjective>, so choose $b\in R$ with $\phi^N(b)=a$. Then
$$
\phi^{N+1}(b)=\phi(a)=0,
$$
so $b\in K_{N+1}=K_N$. Therefore $a=\phi^N(b)=0$.
Thus the <kernel of a ring homomorphism> $\phi$ is zero. This proves that a <surjective endomorphism of a Noetherian ring is an automorphism>, and in particular
$$
\boxed{\phi\text{ is injective}.}
$$
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