 Solution

To prove replacement failure at V omega plus omega, put $M=V_{\omega +\omega }$. The set $\omega$ belongs to $M$, and for every $n\in \omega$ the ordinal $\omega +n$ also belongs to $M$. The relation saying that $y$ is obtained by taking $n$ successive ordinal successors of $\omega$ is definable, with parameter $\omega$, and is single valued in $M$.

For clarity this does not assume Replacement inside $M$. For each finite $n$ its finite computation sequence exists in $M$: its entries have rank below $\omega +n+1$, and the finite coding overhead still has rank below $\omega +\omega$. The bounded formula verifying this finite successor chain is absolute between $M$ and the universe. The ordinary finite induction identifying its unique last entry uses the actual finite ordinals, which are the same in $M$. This justifies the internal definable function $n\mapsto \omega +n$ without invoking the axiom being tested.

If Replacement held there, its range $B=\{\omega +n:n\in \omega \}$ would be an element of $M$. But ordinal rank equals the ordinal, soEvery element of $V_{\omega +\omega }$ has strictly smaller rank. This contradiction proves Replacement fails in $V_{\omega +\omega }$. Each individual output is present; the missing object is the set collecting all outputs.