Given any number \(n\) you can write it as an addition of \(n\) number of \(1\)’s:

\[n = 1 + 1 + \cdots + 1 (n\text{ times})\]

Therefore, we have:

\[n\times 0 = (1 + 1 + \cdots + 1)\times 0\]

By distributivity of multiplication over addition, we have:

\[n\times 0 = 1\times 0 + 1\times 0 + \cdots + 1\times 0\]

By the multiplicative identity, we have:

\[n\times 0 = 0 + 0 + \cdots + 0\]

By the additive identity property, we have:

\[n\times 0 = 0\]