As stated, the inequality indeed fails. But the final estimate (11.27) in the book is essentially correct. From the product rule the derivative

$$ (\frac{d}{dt})^n (t^z (1 - \psi(t\tau)) \chi(t) $$

contains three types of terms:

1. A term where all derivatives fall on the weight $t^z$. This gives the first term in your expansion.

2. Terms, excluding the one above, where no derivatives hit $\psi$.

3. Terms where some number of derivatives hit $\psi$.

The first term you estimate exactly as the authors do, which gives you

$$ \leq \int_{1/\tau}^\infty \prod (z-k) t^{\Re z - N}~\mathrm{d}t = \prod(z-k)\cdot \frac{1}{\Re z - N + 1} \cdot \tau^{N -1 -\Re z} $$

assuming, as they did, that $\Re z - N < -1$.

For the second types of terms, note that since $\chi$ is support on $[-1,1]$ this means that $|t^z| < 1$ on its support. So every single term of type two is estimated by

$$ \leq \sup_{k \in \{1, \ldots, N\}} \sup_{t \in [0,1]} | \chi^{(k)}(t) | $$

in the context of the estimates we will take this to just be a constant depending on $N$. So all terms of type two together is estimated by

$$ \leq C_N$$

We are left with terms of type three. Since a derivative hits $\psi$, the terms of type three are supported in the region $$ t \in [1/\tau, 2/\tau ] $$ which has width $1/\tau$. In this region $t \approx 1/\tau$. So a typical term can be controlled by

$$ \leq C\cdot \frac{1}{\tau} \cdot \prod(z-k) \cdot \frac{1}{\tau}^{\Re z - N_1} \cdot \tau^{1 + N_2} \cdot |\psi^{(1+N_2)}| \cdot |\chi^{(N_3)} $$

where $1 + N_1 + N_2 + N_3 = N$. We can simplify a little bit putting the $L^\infty$ norms of derivative of $\psi$ and $\chi$ into a constant

$$ \leq C_N \prod_{k = 0}^{N_1 - 1}(z - k) \tau^{N_1 + N_2 - \Re z} $$

since we are interested in the $\tau \gg 1$ case

$$ \leq C_N \prod_{k = 0}^{N_1 - 1}(z-k) \tau^{N - 1 - \Re z} $$

So up to some factors of products of the form $z(z-1)\ldots (z-N_1)$ the estimate (11.27) is correct. Since I haven't read the rest of the chapters to see why the authors chose to keep track of the $z$ dependence, I don't know how much this changes the exposition.

As stated, the inequality indeed fails. But the final estimate (11.27) in the book is essentially correct. From the product rule the derivative

$$ (\frac{d}{dt})^n (t^z (1 - \psi(t\tau)) \chi(t) $$

contains three types of terms:

1. A term where all derivatives fall on the weight $t^z$. This gives the first term in your expansion.

2. Terms, excluding the one above, where no derivatives hit $\psi$.

3. Terms where some number of derivatives hit $\psi$.

The first term you estimate exactly as the authors do, which gives you

$$ \leq \int_{1/\tau}^\infty \prod (z-k) t^{\Re z - N}~\mathrm{d}t = \prod(z-k)\cdot \frac{1}{\Re z - N + 1} \cdot \tau^{N -1 -\Re z} $$

assuming, as they did, that $\Re z - N < -1$.

For the second types of terms, note that since $\chi$ is support on $[-1,1]$ this means that $|t^z| < 1$ on its support. So every single term of type two is estimated by

$$ \leq \sup_{k \in \{1, \ldots, N\}} \sup_{t \in [0,1]} | \chi^{(k)}(t) | $$

in the context of the estimates we will take this to just be a constant depending on $N$. So all terms of type two together is estimated by

$$ \leq C_N$$

We are left with terms of type three. Since a derivative hits $\psi$, the terms of type three are supported in the region $$ t \in [1/\tau, 2/\tau ] $$ which has width $1/\tau$. In this region $t \approx 1/\tau$. So a typical term can be controlled by

$$ \leq C\cdot \frac{1}{\tau} \cdot \prod(z-k) \cdot \frac{1}{\tau}^{\Re z - N_1} \cdot \tau^{1 + N_2} \cdot |\psi^{(1+N_2)}|_\infty \cdot |\chi^{(N_3)}|_\infty $$

where $1 + N_1 + N_2 + N_3 = N$. We can simplify a little bit putting the $L^\infty$ norms of derivative of $\psi$ and $\chi$ into a constant

$$ \leq C_N \prod_{k = 0}^{N_1 - 1}(z - k) \cdot \tau^{N_1 + N_2 - \Re z} $$

since we are interested in the $\tau \gg 1$ case

$$ \leq C_N \prod_{k = 0}^{N_1 - 1}(z-k) \cdot \tau^{N - 1 - \Re z} $$

So up to some factors of products of the form $z(z-1)\ldots (z-N_1)$ the estimate (11.27) is correct. Since I haven't read the rest of the chapters to see why the authors chose to keep track of the $z$ dependence, I don't know how much this changes the exposition.