Question:

<b>Taylor Series Recognition</b><br> $f(x)=x^n$, $f^{(k)}(x)=n(n-1)\cdots(n-k+1)x^{n-k}$, so $f^{(k)}(1)=\dfrac{n!}{(n-k)!}$ for $k\leq n$ and 0 for $k>n$.<br> The sum is $\sum_{k=0}^{n}(-1)^k\dfrac{f^{(k)}(1)}{k!} = \sum_{k=0}^{n}(-1)^k\dfrac{n!}{k!(n-k)!} = \sum_{k=0}^n(-1)^k\binom{n}{k}$.<br> By binomial theorem: $\sum_{k=0}^n(-1)^k\binom{n}{k} = (1-1)^n = 0^n = 0$ (for $n\geq 1$).<br> <b>Answer: 0</b> for all $n\geq 1$.<br> <b>Key concept:</b> Recognise the sum as the binomial expansion of $(1-1)^n=0$.<br> <b>Trap:</b> Computing each derivative separately instead of recognising the binomial pattern.