[proofplan] We prove equality of arithmetic functions by evaluating both sides at an arbitrary positive integer $n \in \mathbb{N}$. The first identity follows by expanding the definition of Dirichlet convolution for the two constant-one functions. The second identity follows by expanding the same definition for $\operatorname{id}_k$ and $\mathbf{1}$, then using $\operatorname{id}_k(d)=d^k$. [/proofplan] [step:Expand the convolution of the two constant-one functions] Fix $n \in \mathbb{N}$. By the definition of Dirichlet convolution applied to the arithmetic functions $\mathbf{1}: \mathbb{N} \to \mathbb{C}$ and $\mathbf{1}: \mathbb{N} \to \mathbb{C}$, \begin{align*} (\mathbf{1} * \mathbf{1})(n) &= \sum_{d \mid n} \mathbf{1}(d)\mathbf{1}\left(\frac{n}{d}\right) \\ &= \sum_{d \mid n} 1 \cdot 1 \\ &= \sum_{d \mid n} 1 \\ &= \tau(n). \end{align*} Since this holds for every $n \in \mathbb{N}$, the arithmetic functions are equal: \begin{align*} \tau = \mathbf{1} * \mathbf{1}. \end{align*} [/step] [step:Expand the convolution of the power function with the constant-one function] Again fix $n \in \mathbb{N}$. By the definition of Dirichlet convolution applied to $\operatorname{id}_k: \mathbb{N} \to \mathbb{C}$ and $\mathbf{1}: \mathbb{N} \to \mathbb{C}$, \begin{align*} (\operatorname{id}_k * \mathbf{1})(n) &= \sum_{d \mid n} \operatorname{id}_k(d)\mathbf{1}\left(\frac{n}{d}\right) \\ &= \sum_{d \mid n} d^k \cdot 1 \\ &= \sum_{d \mid n} d^k \\ &= \sigma_k(n). \end{align*} Since this holds for every $n \in \mathbb{N}$, the arithmetic functions are equal: \begin{align*} \sigma_k = \operatorname{id}_k * \mathbf{1}. \end{align*} This proves both asserted convolution identities. [/step]