$ (1*...*1)(n) =\# \{(m_1,...,m_k)\in (\mathbb{N}\setminus \{0\})^k : m_1...m_k = n\} $

We call an arithmetic function any element of $\mathcal{F} (\mathbb{N}\setminus \{0\} , \mathbb{C})$. We endow $\mathcal{F} (\mathbb{N} \setminus \{0\}, \mathbb{C})$ binary operation convolution and defined by $(f*g) (n): = \sum_{d|n}{f (d) g (n/d)}$ that is commutative and associative.

We consider the function $\tau_k \in \mathcal{F} (\mathbb{N}\setminus \{0\} , \mathbb{C})$ defined by $$\tau_k = \underbrace{1*...*1}_{k \text{ times}}$$ where $1\in \mathcal{F} (\mathbb{N}\setminus \{0\}, \mathbb{C})$ denotes the constant function being equal to $1$. When $n \in \mathbb{N}\setminus \{0\} $, I see if $k=1$ (resp, $2$) then $\tau_1(n)=1$ (resp, $\tau_2(n)= \# \{\text{ divisor of n}\}$)

My problem why, for $k\geq 2$: $$ \tau_k(n) =\# \{(m_1,...,m_k)\in (\mathbb{N}\setminus \{0\})^k : m_1...m_k = n\} $$

Assume $\tau_k(n) = \# \{(m_1 , \dots , m_k) \in \mathbb{Z}_+^k | m_1 \dots m_k = n \} $

Then

\begin{align*} (\tau_k \ast 1 )(n) & = \sum_{d |n} \# \{(m_1 , \dots , m_{k}) \in \mathbb{Z}_+^{k} | m_1 \dots m_{k} = \frac{n}{d} \} \\ & = \sum_{d |n} \# \{(m_1 , \dots , m_{k}) \in \mathbb{Z}_+^{k} | m_1 \dots m_{k}d = n \} \\ & = \# \{(m_1 , \dots , m_{k},d) \in \mathbb{Z}_+^{k+1} | m_1 \dots m_{k}d = n \} \\ & = \tau_{k+1}(n) \end{align*}