[proofplan] We compute the divisor sum by using the prime factorization of $n$. The von Mangoldt function vanishes on every divisor except prime powers, so the sum over all divisors reduces to a sum over the prime-power divisors appearing in the factorization of $n$. Each prime $p_i$ contributes exactly $a_i$ copies of $\log p_i$, which gives $\log(p_1^{a_1}\cdots p_r^{a_r})=\log n$. [/proofplan] [step:Separate the case $n=1$ from the prime factorization case] For $n=1$, the only positive divisor of $1$ is $1$. Since $1$ is not a prime power, $\Lambda(1)=0$. Hence \begin{align*} \sum_{d \mid 1}\Lambda(d)=\Lambda(1)=0=\log 1. \end{align*} It remains to consider $n \geq 2$. [/step] [step:Write $n$ as a product of prime powers] Let $n \geq 2$. Write the prime factorization of $n$ as \begin{align*} n=\prod_{i=1}^{r} p_i^{a_i}, \end{align*} where $r \in \mathbb{N}$, the numbers $p_1,\dots,p_r$ are distinct primes, and $a_1,\dots,a_r \in \mathbb{N}$. A positive divisor $d$ of $n$ has the form \begin{align*} d=\prod_{i=1}^{r} p_i^{b_i} \end{align*} for uniquely determined integers $b_i$ satisfying $0 \leq b_i \leq a_i$. [/step] [step:Identify exactly which divisors have nonzero von Mangoldt value] By definition of $\Lambda$, a divisor $d \mid n$ contributes nontrivially to the sum only when $d$ is a positive power of a single prime. Since every prime dividing $d$ must be among $p_1,\dots,p_r$, the contributing divisors are exactly \begin{align*} p_i^j \quad \text{with } 1 \leq i \leq r \text{ and } 1 \leq j \leq a_i. \end{align*} For each such divisor, the definition of the von Mangoldt function gives \begin{align*} \Lambda(p_i^j)=\log p_i. \end{align*} Every other divisor has at least two distinct prime factors, or is equal to $1$, and therefore has von Mangoldt value $0$. [/step] [step:Sum the surviving contributions and recover $\log n$] Using the preceding identification of the nonzero terms in the divisor sum, we obtain \begin{align*} \sum_{d \mid n}\Lambda(d) &=\sum_{i=1}^{r}\sum_{j=1}^{a_i}\Lambda(p_i^j)\\ &=\sum_{i=1}^{r}\sum_{j=1}^{a_i}\log p_i\\ &=\sum_{i=1}^{r} a_i \log p_i. \end{align*} The logarithm turns products into sums and powers into scalar multiples, so \begin{align*} \log n &=\log\left(\prod_{i=1}^{r} p_i^{a_i}\right)\\ &=\sum_{i=1}^{r} \log(p_i^{a_i})\\ &=\sum_{i=1}^{r} a_i \log p_i. \end{align*} Therefore \begin{align*} \sum_{d \mid n}\Lambda(d)=\log n. \end{align*} Since this holds for every $n \in \mathbb{N}$, the arithmetic functions are equal pointwise: \begin{align*} \log = \Lambda * \mathbb{1}. \end{align*} [/step]