Setup:
Consider a renewal process where $Y_i\overset{iid}{\sim} p$ for $i\in \mathbb{Z}_{+}$ and $p$ is a probability distribution
$$p:\mathbb{Z}_{+}\rightarrow [0,1].$$
If we define $Z_k=\sum_{i=0}^{k}Y_i$ and consider
$$u(n)=\sum_{k=0}^{\infty}\mathbb{P}(Z_k=n).$$
The book states that this is true:
$$u(n)=1 + u\circledast p(n).$$
I keep getting something different:
$$u(n)=\mathbb{P}(Z_0=n) + \sum_{k=1}^{\infty}\sum_{j=0}^{n-1}\mathbb{P}(Z_k=n, Z_{k-1}=j)=$$ $$p(n)+\sum_{k=1}^{\infty}\sum_{j=0}^{n-1}\mathbb{P}(Y_k=n-j, Z_{k-1}=j)=p(n)+\sum_{k=1}^{\infty}\sum_{j=0}^{n-1}\mathbb{P}(Y_k=n-j)\mathbb{P}( Z_{k-1}=j)=$$ $$=p(n)+\sum_{k=1}^{\infty}\sum_{j=0}^{n-1}p(n-j)p^{\circledast(k-1)}(j)=p(n)+\sum_{j=0}^{n-1}p(n-j)u(j)=p(n)+u\circledast p(n).$$
The last step follows because $p(0)=0$ by convention. I don't believe the $p(n)=1$ so I don't know where my mistake lies.
Secondly, if this identity holds and we define $u(z)=\sum_{k=0}^{\infty}u(k)z^k$ and $P(z)=\sum_{k=0}^{\infty}p(k)z^k$, then we get
$$u(z)=(1-P(z))^{-1} \hspace{2mm} \text{for } |z|<1.$$
I get that I can multiply $z^n$ to both sides and get
$$u(n)z^n=z^n + u\circledast p(n)z^n.$$
Then I can take the sum:
$$u(z)=\sum_{k=0}^{\infty}z^{k} + u(z)P(z).$$
From here I run into the issue of solving for $u(Z)$ if the first series to the right of the equality was 1 then I can easily get the result but I am not guaranteed that. What mistake have I made?
Edit: Where $\circledast$ is a convolution of two functions.