I'm trying to understand section (VI.4) on smooth base change in Milne's Étale Cohomology. He defines a morphism $g \colon Y \longrightarrow X$ to be $n$-acyclic (I only care about the case $n \geq 0$) if for any scheme $X'$ étale and of finite type over $X$ and any appropriate torsion sheaf $\mathcal{F}$ on $X'$, $H^i(X', \mathcal{F}) \longrightarrow H^i(Y', g'^*\mathcal{F})$ is bijective for all $0 \leq i \leq n$ and injective for $i=n+1$. Here $g' \colon Y' = Y \times_X X' \longrightarrow X'$ is the base change of $g$.
It is stated that this is equivalent to the condition $(*)$ that for any $X'$ and $\mathcal{F}$ as above, the map $\mathcal{F} \longrightarrow g'_*g'^*\mathcal{F}$ is an isomorphism and $R^jg'_*(g'^*\mathcal{F}) = 0$ for $1 \leq j \leq n$.
I see why the definition implies the condition $(*)$. However, I'm a bit confused about the other direction. I don't have a lot of experience with spectral sequences yet, I think that's where the problem lies.
The Leray spectral sequence for $g'$ reads $H^i(X',R^jg'_*(g'^*\mathcal{F})) \Longrightarrow H^{i+j}(Y',g'^*\mathcal{F})$. Now, using $(*)$, we find that, since the differentials on the $r$-th page have grading $(r,1-r)$, the entries at $(i,0)$ with $0 \leq i \leq n+1$ in the spectral sequence never have any non-zero arrows coming in our out of them on any page. Hence, they are already stable from the second page on, i.e. $H^i(X', \underbrace{g'_*g'^*\mathcal{F}}_{\mathcal{F}}) \cong H^i(Y',g'^*\mathcal{F})$ for $0 \leq i \leq n+1$. But this is not what we want for $i = n+1$.
It would be nice if someone could point out what I'm misunderstanding.
$\newcommand{\cF}{\mathcal F}$On the second page, out of $H^i(X',R^jg_*'(g'_*\cF))$ for $i+j=n+1$, you have two possible nonzero terms: $$ E^{n+1,0}_2 = H^{n+1}(X',g_*'(g'_*\cF)),\quad E^{0,n+1}_2=H^{0}(X',R^{n+1}g_*'(g'_*\cF)). $$ (I think you might be forgetting about the second one) The first one has no more nonero arrows in any pages, so like you say $E^{n+1,0}_2 \cong E^{n+1,0}_\infty$, but we don't really know what $E^{0,n+1}_\infty$ looks like, and it might be nonzero.
Once the spectral sequence has converged, what it tells us is that $H^{n+1}(Y',g'^*\cF)$ has a filtration whose graded pieces are $E^{p,q}_\infty$ for $p+q=n+1$. In this case, there are at most two nonzero graded pieces, so we just have a short exact sequence: $$ 0\to E^{n+1,0}_\infty \to H^{n+1}(Y',g'^*\cF) \to E^{0,n+1}_\infty\to 0. $$ Since $E^{n+1,0}_\infty \cong H^{n+1}(X',g_*'(g'^*\cF))$, there you have your injection, but since $E^{0,n+1}_\infty$ might be nonzero, it might not be surjective. The map in the spectral sequence comes from the adjunction map $\cF\to g'_*g'^*\cF$... I'm not sure what is the best way to see this. I would say that it's to see that there is a map of double complexes from the complex computing $H^i(X',g_*'(g'^*\cF))$ (seen as a double complex with just one nonzero row or column) to the one the Leray spectral sequence comes from, which will induce the map in the short exact sequence above.
The exact same reasoning can be repackaged with no spectral sequences, just a short exact sequence. I'm going to write it down since I personally find it helpful to understand spectral sequences when I can explain what's happening with long exact sequences.
Let $I^\bullet$ be an injective resolution of $Rg'_*(g'^*(\cF))$. Then, there is a short exact sequence $$ 0\longrightarrow \tau_{\le n}I^\bullet \longrightarrow I^\bullet \longrightarrow \tau_{> n}I^\bullet \longrightarrow 0. $$ $\tau_{\le n}I^\bullet$ is the truncation where we replace $I^n$ by $\ker d\subseteq I^n$ and $(\tau_{\le n}I^\bullet)^i=0$ for $i>n$., and $\tau_{>n}$ is just the quotient.
Assuming that $R^ig'_*(g'^*(\cF))=0$ for $1<i\le n$ exactly means that $g'_*(g'^*(\cF))\cong \tau_{\le n}Rg'_*(g'^*(\cF))$. Also, $\tau_{>n}I^\bullet$'s cohomology is $0$ up to degree $n$. So, the short exact sequence gives us a long exact sequence taking $H^*(X',\bullet)$, which has a bunch of zeroes, so we split it into: $$ i<n\colon \quad 0\to H^i(X',g'_*(g'^*(\cF))) \to H^i(Y',\cF) \to 0. $$ $$ 0\to H^{n+1}(X',g'_*(g'^*(\cF))) \to H^{n+1}(Y',\cF) \to H^{n+1}(X',\tau_{>n}Rg'_*(g'^*(\cF)))\to \cdots $$