= Solution
For a <profinite abelian group> $G$, its <Iwasawa algebra> is the completed <group algebra>
$$
\boxed{\Lambda(G)=\varprojlim_U\mathbb Z_p[G/U]=\varprojlim_{U,r}(\mathbb Z/p^r\mathbb Z)[G/U],}
$$
where $U$ runs over open <subgroups>. The transition map sums the coefficients over the fibers of $G/V\to G/U$ when $V\subset U$. Consequently an element gives compatible values $\mu(gU)\in\mathbb Z_p$ on <cosets>, equivalently a finitely additive $\mathbb Z_p$-valued <p-adic measure> on the <clopen sets> of $G$. Conversely these values specify every finite-quotient coefficient. Integration of a <continuous function> with values in $\mathbb Z_p$ is obtained by uniform approximation by locally constant functions; the integrals converge because all measure values have $p$-adic absolute value at most $1$. Multiplication in the <Iwasawa algebra> becomes <convolution of p-adic measures>. This is its <measure realization of an Iwasawa algebra>.
The <Mahler theorem> states that each <continuous function> $g:\mathbb Z_p\to\mathbb Z_p$ has the unique expansion converging in the sense of <uniform convergence>
$$
g(x)=\sum_{n\geq0}b_n\binom{x}{n},\qquad b_n\in\mathbb Z_p,\quad b_n\longrightarrow0.
$$
Its <Mahler coefficients> are
$$
b_n=\Delta^ng(0)=\sum_{j=0}^{n}(-1)^{n-j}\binom njg(j),
$$
and $\|g\|_\infty=\sup_n|b_n|_p$. The same assertion holds for <continuous functions> with values in $\mathbb Q_p$ with coefficients in $\mathbb Q_p$.
Given $f(T)=\sum_{n\geq0}a_nT^n\in R$, define the <bounded linear functional>
$$
L_f(g)=\sum_{n\geq0}a_nb_n.
$$
The sum converges because $a_n\in\mathbb Z_p$ and $b_n\to0$. Its values on <indicator functions> of <clopen sets> give a <p-adic measure>, denoted $\lambda(f)$, with
$$
\int_{\mathbb Z_p}\binom{x}{n}\,d\lambda(f)=a_n.
$$
Conversely, a <p-adic measure> determines these coefficients $a_n\in\mathbb Z_p$, and continuity allows termwise integration of every <Mahler expansion>. Thus the inverse is
$$
\mu\longmapsto\sum_{n\geq0}\left(\int\binom{x}{n}\,d\mu\right)T^n,
$$
which is the <Amice transform>. These two constructions are inverse, proving the \b[canonical bijection $\lambda:R\simeq\Lambda(\mathbb Z_p)$]. In fact it is an <isomorphism> of <rings>: the identity $\binom{x+y}{n}=\sum_{i+j=n}\binom{x}{i}\binom{y}{j}$ identifies <convolution of p-adic measures> with multiplication of <formal power series>. For example, $\lambda(1)=\delta_0$ and $\lambda(1+T)=\delta_1$, where $\delta_a$ is the <Dirac measure> at $a$.
For the moment identity, let $\left\{{k\atop n}\right\}$ denote a <Stirling number of the second kind>. The finite <polynomial identity>
$$
x^k=\sum_{n=0}^{k}n!\left\{{k\atop n}\right\}\binom{x}{n}
$$
gives $\int x^k\,d\lambda(f)=\sum_{n=0}^{k}a_n n!\left\{{k\atop n}\right\}$. On the other hand, $D=(1+T)d/dT$ satisfies $D^k(1+T)^x=x^k(1+T)^x$ for every nonnegative integer $x$. Expanding $(1+T)^x$ and evaluating at $T=0$ yields
$$
x^k=\sum_{n=0}^{k}\binom{x}{n}(D^kT^n)(0).
$$
Comparison in the <basis> of binomial <polynomials> gives $(D^kT^n)(0)=n!\left\{{k\atop n}\right\}$. Terms with $n>k$ have zero constant term after at most $k$ differentiations, so no infinite-series interchange is needed. Therefore
$$
\boxed{\int_{\mathbb Z_p}x^k\,d\lambda(f)=(D^kf)(0)\qquad(k\geq0).}
$$
The case $k=0$ says that the total mass is $f(0)$.
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