Let $p$ be a prime number. There is a well known product formula for $\log(1+X)$ as a power series in $\mathbb{Z}_p[[X]]$ . Namely letting $\Phi_n(X)=(1+X)^{p^n}-1$ and $Q_n(X)=\Phi_{n+1}(X)/p\Phi_n(X)$, then we have$$\log(1+X)=X\prod_{n\geq1}Q_n(X).$$.
How can I prove this?
Pdf I'm reading reads both sides have exactly the same zeros, and up to a constant this completely determines a power series in $\mathbb{Z}_p[[X]]$ by the weierstrass approximation theorem. But I cannot figure out how and which statement we apply to this problem.
Another way or reference is also appreciated, thank you in advance.
For any $a\in p^2\Bbb{Z}_p$, write $1+a=\exp(b)$, we have $$ \lim_{k\to \infty} \frac{(1+a)^{p^k}-1}{p^k}=\lim_{k\to \infty} \frac{\exp(p^k b)-1}{p^k} = \lim_{k\to \infty} b + O(p^k) = b= \log(1+a)$$ So the infinite product and $\log(1+X)$ agree as functions $p^2\Bbb{Z}_p\to \Bbb{Z}_p$, which is a good start.