Let $k$ be a commutative ring with $1$ and $A$ a commutative unital $k$-algebra ($k$ and $A$ are assumed to be associative). Denote by $(C_\bullet(A),b)$ the Hochschild chain complex of $A$ and let $$ C(A) = \bigoplus_{n=0}^{\infty} C_n(A). $$ Let $S_{p,q}$ denote the set of $(p,q)$-shuffles in the symmetric group $S_{p+q}$, that is, a permutation $\sigma\in S_{p+q}$ belongs to $S_{p,q}$ if and only if $$ \sigma(1)<\sigma(2)<\cdots <\sigma(p) \quad \text{and} \quad \sigma(p+1)<\cdots<\sigma(p+q). $$ The shuffle product on $C(A)$ is definen on homogeneous elements $\alpha=a_0\otimes a_1\otimes\cdots\otimes a_p\in C_p(A)$ and $\beta=a_0'\otimes a_{p+1}\otimes \cdots \otimes a_{p+q}\in C_q(A)$ by $$ \alpha \times \beta = \sum_{\sigma\in S_{p,q}} \mbox{sign}(\sigma) a_0a_0'\otimes a_{\sigma^{-1}(1)}\otimes a_{\sigma^{-1}(2)}\otimes\cdots\otimes a_{\sigma^{-1}(p+q)}. $$
I proved directly from this definition that the boundary $b$ of the Hochschil complex is a graded derivation for the shuffle product, that is, if $\alpha\in C_p(A)$ and $\beta\in C_q(A)$, then $$ b(\alpha\times \beta) = b(\alpha)\times \beta + (-1)^p \alpha\times b(\beta). $$ Also, if for $\alpha$ as above we define $$ \Delta(\alpha) = \sum_{i=1}^{p} a_0\otimes a_1\otimes \cdots a_i\otimes 1\otimes a_{i+1}\otimes \cdots \otimes a_p, $$ this gives $C(A)$ the structure of a coalgebra, and I proved (again, directly from these definitions) that $b$ is a coderivation for $\Delta$. But my proofs are painfully tedious. So I was wondering if it is possible to give a more conceptual proof of the above facts.
Maybe there are some theoretical facts that I don't know that allow me to write a more concise and elegant proof.
Thank you in advance!