The inverse of one plus the generator in the truncated mod-two polynomial ring

Every element of $R_m$ has a unique representative of degree at most $m$ by [F1], and $t^{m+1}=0$; in particular $1,t,\dots ,t^m$ form a basis and no class has two such representatives. The element $1+t$ is a unit with the displayed finite inverse: since $m\ge 1$, in characteristic two $(1+t)(1+t+\cdots +t^m)=1+t^{m+1}=1+0=1$ in $R_m$, so $1+t+\cdots +t^m$ is an inverse of $1+t$ and hence $1+t$ is a unit.

Write $m={\sum }_{j\in B}{2}^j$ with $B$ the finite set of binary digit positions, so $B\mathrm{\ne }\mathrm{\varnothing }$ because $m\ge 1$. Choose $M\ge \max B$ with $2^{M+1}>m$. In ${\mathbb{F}}_2[t]$ the identity $(1+t{)}^{2^j}=1+t^{2^j}$ holds for every $j\ge 0$: it is trivial for $j=0$, and from $(u+v{)}^2=u^2+2uv+v^2=u^2+v^2$ in characteristic two, $(1+t{)}^{2^{j+1}}=((1+t{)}^{2^j}{)}^2=(1+t^{2^j}{)}^2=1+t^{2^{j+1}}$. Multiplying the identities for $j\in B$ gives $(1+t{)}^m={\prod }_{j\in B}(1+t^{2^j})$, and iterating ${\prod }_{j=0}^N(1+t^{2^j})=1+t+\cdots +t^{2^{N+1}-1}$ gives, in $R_m$, $(1+t{)}^{m+1}=(1+t){\prod }_{j\in B}(1+t^{2^j}),{\prod }_{j=0}^M(1+t^{2^j})=1+t+\cdots +t^{2^{M+1}-1}\mathrm{.}$

Put $P_m:={\prod }_{j\mathrm{\notin }B,0\le j\le M}(1+t^{2^j})$, a finite product in $R_m$. Expanding the product over all subsets $A\subseteq \{0,\dots ,M\}\setminus B$, each subset contributes $t^{\sigma (A)}$ with $\sigma (A)={\sum }_{j\in A}{2}^j$, and distinct subsets have distinct sums by uniqueness of binary expansion; all other coefficients are $0$. Since $t^k=0$ in $R_m$ for every $k\ge m+1$ by [F1], only the subsets with $\sigma (A)\le m$ contribute, and such a sum has binary support inside $\{0,\dots ,M\}$ and disjoint from $B$, i.e. $\sigma (A)\wedge m=0$, so $\sigma (A)\in S_m$. Conversely every $i\in S_m$ satisfies $i\le m<2^{M+1}$, so its binary expansion involves only digits $j\le M$ and, since $i\wedge m=0$, no digit of $B$; the subset $A=\{j:2^j\text{ occurs in }i\}$ is admissible and $i=\sigma (A)$. Therefore $P_m={\sum }_{i\in S_m}{t}^i\text{in }{R}_m\mathrm{.}$

In $R_m$ one computes, using step 1.2 and the Frobenius identities, $(1+t{)}^{m+1}{P}_m=(1+t){\prod }_{j=0}^M(1+t^{2^j})=(1+t)(1+t+\cdots +t^{2^{M+1}-1})=1+t^{2^{M+1}}=1,$ the last equality because $2^{M+1}>m$ forces $t^{2^{M+1}}=0$ in $R_m$ by [F1]. Hence $P_m$ is a two-sided inverse of $(1+t{)}^{m+1}$ in the commutative ring $R_m$, so $(1+t{)}^{-(m+1)}=P_m={\sum }_{i\in S_m}{t}^i$ by step 2.1.

For the binomial-coefficient description apply [F2] over the commutative ring ${\mathbb{F}}_2$ with $\lambda =1$ and $j=m+1$: in ${\mathbb{F}}_2\mathrm{⟦}x\mathrm{⟧}$ one has $(1-x{)}^{-(m+1)}={\sum }_{i\ge 0}(\left(\genfrac{}{}{0px}{}{i+m}{m}\right)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2)x^i={\sum }_{i\ge 0}(\left(\genfrac{}{}{0px}{}{m+i}{i}\right)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2)x^i$, the second equality by the symmetry clause of [F2], where $1-x=1+x$ because $1+1=0$ in ${\mathbb{F}}_2$. By [F3] the coefficientwise truncation map $\phi :{\mathbb{F}}_2\mathrm{⟦}x\mathrm{⟧}\to R_m$, ${\sum }_i{a}_i{x}^i\mapsto {\sum }_{i=0}^m{a}_i{t}^i$, is a surjective ring homomorphism: addition is coefficientwise, and in the Cauchy product the coefficient of $x^k$ depends only on the coefficients of degree at most $k$, so truncation at degree $m$ commutes with products in $R_m$, where $t^{m+1}=0$. Since $\phi (1-x)=1+t$ and ring homomorphisms carry inverses of units to inverses of units, $\phi ((1-x{)}^{-(m+1)})=(1+t{)}^{-(m+1)}={\sum }_{i=0}^m(\left(\genfrac{}{}{0px}{}{m+i}{i}\right)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2)t^i\mathrm{.}$ Comparing coefficients with step 3.1 gives: the coefficient of $t^i$ in $(1+t{)}^{-(m+1)}$ equals $\left(\genfrac{}{}{0px}{}{m+i}{i}\right)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$ for every $0\le i\le m$, and it equals $1$ exactly for the $i\in S_m$ by the formula of step 3.1.

The set $S_m$ contains $0$ and is finite, so $d(m)=\max S_m$ is defined; by steps 3.1 and 4.1 the coefficient of $t^{d(m)}$ is $1$, while every coefficient of degree $i>d(m)$ with $i\le m$ is $0$ because such $i\mathrm{\notin }{S}_m$, and degrees above $m$ vanish in $R_m$. Hence the highest power occurring with nonzero coefficient is exactly $t^{d(m)}$. If $m+1=2^q$ is a power of two, then $q\ge 1$, $m=2^q-1={\sum }_{j=0}^{q-1}{2}^j$, and every $i$ with $1\le i\le m$ has some binary digit at a position $j\le q-1$, hence satisfies $i\wedge m\mathrm{\ne }0$; therefore $S_m=\{0\}$ and the inverse is $1$, consistently with $(1+t{)}^{m+1}=(1+t{)}^{2^q}=1+t^{2^q}=1$ in $R_m$.