For $p$ an odd prime and $i\ge 1$ integer there exists $\nu \in \mathbb{Z}$ such that : $(1+p)^{{p}^{i}}=1+\nu p^{i+1}$

Question

With $\gcd{(\nu,p)}=1$.

I try to see it in $\pmod {p^{i+2}}$ and I obtained $1+p^{i+1}$ but the equality does not hold anymore in $\mathbb{Z}$

Thanks in advance !

Answer 1

Induct on $i$.

For $i=0$, $(1+p)^1 = 1 + \nu_0 p^1$ with $\nu_0 = 1$.

If $(1+p)^{p^i} = 1 + \nu_i p^{i+1}$, then $$ (1+p)^{p^{i+1}} = (1 + \nu_i p^{i+1})^p = 1 + \nu_i p^{i+2} + \sum_{j=2}^p {p \choose j} \nu_i^j p^{j(i+1)} = 1 + \nu_{i+1} p^{i+2}$$ where $$ \nu_{i+1} = \nu_i + \sum_{j=2}^p {p \choose j} \nu_i^j p^{(j-1)i+j-2} $$ and note that ${p \choose j} p^{(j-1)i+j-2}$ is divisible by $p$.