Let $n = 2^\alpha \cdot 3^\beta \cdot 5^\gamma \cdot 7^\delta \cdot s$, where $s$ is not divisible by $2$, $3$, $5$ or $7$; then $5n = 2^\alpha \cdot 3^\beta \cdot 5^{\gamma+1} \cdot 7^\delta \cdot s$, $6n = 2^{\alpha+1} \cdot 3^{\beta+1} \cdot 5^\gamma \cdot 7^\delta \cdot s$ and $7n = 2^\alpha \cdot 3^\beta \cdot 5^\gamma \cdot 7^{\delta+1} \cdot s$. Consequently:
* For $\sqrt[5]{5n}$ to be an integer, $\alpha$, $\beta$, $\gamma + 1$ and $\delta$ must be divisible by $5$;
* For $\sqrt[6]{6n}$ to be an integer, $\alpha + 1$, $\beta + 1$, $\gamma$ and $\delta$ must be divisible by $6$;
* For $\sqrt[7]{7n}$ to be an integer, $\alpha$, $\beta$, $\gamma$ and $\delta + 1$ must be divisible by $7$.
Hence $\alpha$ and $\beta$ must be divisible by $35$, $\gamma$ must be divisible by $42$ and $\delta$ must be divisible by $30$. The least suitable value for $\alpha$ and $\beta$ is $35$ since $35$ is the least positive multiple of $35$ and the next integer is divisible by $6$. Studying the multiples of $42$ and $30$ similarly shows that the least suitable value for $\gamma$ is $84$ and the least suitable value for $\delta$ is $90$. For the least suitable value for $n$, take $s = 1$. Hence the desired number is $2^{35} \cdot 3^{35} \cdot 5^{84} \cdot 7^{90}$.