Find the n -th derivative of y = x^n

Find the $n$-th derivative of $y = x^{n}$.

Step 1: Differentiate repeatedly

\[ y' = nx^{\,n-1}, \qquad y'' = n\bigl(n-1\bigr)x^{\,n-2}, \qquad y''' = n\bigl(n-1\bigr)\bigl(n-2\bigr)x^{\,n-3}. \]

The pattern is clear: after $k$ steps, \[y^{(k)} = n\bigl(n-1\bigr)\cdots\bigl(n-k+1\bigr)x^{\,n-k} = \frac{n!}{\bigl(n-k\bigr)!}\,x^{\,n-k}, \qquad 0\le k\le n. \tag{1}\]

Step 2: Induction

The base case $k = 1$ is the power rule. If (1) holds for $k$, differentiating once more multiplies by $n-k$ and lowers the exponent: \[y^{(k+1)} = \frac{n!}{\bigl(n-k\bigr)!}\bigl(n-k\bigr)x^{\,n-k-1} = \frac{n!}{\bigl(n-k-1\bigr)!}\,x^{\,n-(k+1)},\]

which is (1) with $k+1$ in place of $k$.

Step 3: Set $k = n$

The exponent reaches zero and the falling product becomes a full factorial: $y^{(n)} = n\bigl(n-1\bigr)\cdots2\cdot1\cdot x^{0} = n!$.

Step 4: Verification

\[n = 3:\ y = x^{3},\ y' = 3x^{2},\ y'' = 6x,\ y''' = 6 = 3! \;\;\checkmark\]

\[y^{(n+1)} = \bigl(n!\bigr)' = 0, \text{ as it must be for a polynomial of degree } n \;\;\checkmark\]

\[\text{Maclaurin: the coefficient of } x^{n} \text{ in } x^{n} \text{ is } \frac{y^{(n)}(0)}{n!} = \frac{n!}{n!} = 1 \;\;\checkmark\]