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  "body": "\n## Proof\n\n1. For $0 \\le k \\le n$, let $A_k$ be the matrix whose rows $1, \\dots, k$ are $c$ times the rows of $A$, and whose other rows are those of $A$.\n2. Then $A_0 = A$. By the definition of the [[scalar multiple|09V6DDACXR]], every entry of $cA$ is $c$ times the entry of $A$, so $A_n = cA$.\n3. For $1 \\le k \\le n$, the matrix $A_k$ results from $A_{k-1}$ when row $k$ is multiplied by $c$.\n4. By the rule for [[scaling a row|009CY8JXFX]], $\\det A_k = c \\det A_{k-1}$ for $1 \\le k \\le n$.\n5. By induction on $k$, step 4 gives $\\det A_k = c^k \\det A$ for $0 \\le k \\le n$.\n6. For $k = n$, this is $\\det(cA) = c^n \\det A$.\n\n## Idea\n\nMultiplying a matrix by $c$ multiplies each of its $n$ rows by $c$. Each row scaling brings out one factor $c$, so in total we get $c^n$.\n",
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