I am reading Algebraic Number Theory by Neukirch. There is this apparently simple proposition, but the proof goes too fast for me I think.
I understand that we do a proof by induction. If $n=1$, then we have $a\bmod{p}$ and there is a unique integer $0\leq a_0<p$ such that $a\equiv a_0\bmod{p}$.
The problem is in the induction step. The induction hypothesis is that for $n-1$ the proposition holds. So in case of $n$, we want to prove that there exist unique $0\leq b_i<p$ for $i=0,\ldots,n$ such that $$a\equiv b_0+b_1 p+b_2 p^2+\ldots+b_{n-1}p^{n-1}+b_n p^n \bmod{p^{n+1}}.$$ If we consider $a\bmod{p^n}$, by induction hypothesis the thing was unique, so we must have $b_i=a_i$ for all $i=0,\ldots,n-1$. It remains to show that there exists $b_n$ and that it is unique.
Could someone help?
The induction step $n-1$ seems to guarantee you the uniqueness of
$$a\equiv a_0+a_1 p+\ldots+a_{n-2}p^{n-2} \pmod{ p^{n-1}}.$$
Then going back to equality you might need to add any multiple of $p^{n-1}$, that is
$$a= a_0+a_1 p+\ldots+a_{n-2}p^{n-2} +g p^{n-1}.$$
They let $g\equiv a_{n-1}\pmod p$ where $0\leq a_{n-1}<p$ which is unique, and since $g=a_{n-1} +m\cdot p$ for some integer $m$, then subbing in and reducing mod $p^n$ is what they do next,
\begin{align} a&= a_0+a_1 p+\ldots+a_{n-2}p^{n-2} +(a_{n-1}+m\cdot p) p^{n-1}\\ &\equiv a_0+a_1 p+\ldots+a_{n-2}p^{n-2} +a_{n-1} p^{n-1}\pmod {p^n}. \end{align}
Then provided all of $0\leq a_i<p$ the representation should be unique. (That's in the hypothesis of the proposition as well so you get it from the induction step for $0\leq i\leq n-2$, and $a_{n-1}$ was picked as such)