Feels this can be done quicker. Let $P(x) = \textstyle \sum_{0\le i\le d}\alpha_i x^i$. Baris already gave an example for $d=4$, so let's prove $d\ge 5$ is impossible. Assume the contrary, let $r_1,\dots,r_d$ be roots of $P$ and $\alpha_d\ne 0$. First, suppose $\alpha_0=\alpha_1=\alpha_2=1$. Note that by Vieta \[ P\triangleq r_1r_2\cdots r_d = (-1)^d \alpha^{-1}_d\qquad\text{and}\qquad P\sum_{1\le i\le d}\frac{1}{r_i} = (-1)^{d-1} \alpha^{-1}_d \implies \sum_{1\le i\le d}\frac{1}{r_i}=-1. \]Likewise, \[ P\sum_{1\le i<j\le d}\frac{1}{r_i r_j} = (-1)^{d-2}\alpha^{-1}_d \implies \sum_{1\le i<j\le d}\frac{1}{r_ir_j} = 1. \]We now declare a contradiction: observe that \[ \sum_i \frac{1}{r_i^2} = \left(\sum_i \frac{1}{r_i}\right)^2 - 2\sum_{i<j}\frac{1}{r_ir_j} = -1. \]Hence, (exactly) one of $\alpha_0,\alpha_1,\alpha_2$ is different from one. This, in particular, implies $\alpha_d=\alpha_{d-1}=\alpha_{d-2}=1$ as $d\ge 5$. Executing the same logic, \[ \sum_{1\le i\le d}r_i = -1 \qquad\text{and}\qquad \sum_{1\le i<j\le d}r_ir_j = 1 \implies \sum_{1\le i\le d}r_i^2 =-1, \]again a contradiction. Thus, $d\ge 5$ is impossible.

Let's do something even quicker We'll just consider the impossibility of $d\ge 5$ Apply Newton's Inequality for the roots of our polynomial, we obtain $ \frac{a_{n}}{\binom{n}{0}} \frac{a_{n-2}}{\binom{n}{2}} \leq \left(\frac{a_{n-1}}{\binom{n}{1}}\right)^2$ and $ \frac{a_{2}}{\binom{n}{n-2}} \frac{a_{0}}{\binom{n}{n}} \leq \left(\frac{a_{1}}{\binom{n}{n-1}}\right)^2$, where the right side is equivalent to $ \frac{a_{2}}{\binom{n}{2}} \frac{a_{0}}{\binom{n}{0}} \leq \left(\frac{a_{1}}{\binom{n}{1}}\right)^2$. It can be easily seen that in both inequalities, all of the three coefficients cannot be equal to $1$ at the same time, as it gives a contradiction. So, our only non-1 coefficient should be inside both the first and last three coefficients at the same time, giving the desired contradiction.

$P(x) = x^d+ x^{d-1} + ... + x+ 1 + a\cdot x^k = \frac{x^{d+1} -1}{x-1} + a \cdot x^k$ $(x-1)P(x) = x^{d+1} - 1 + a \cdot x^k(x-1) = R(x)$ $R(x)$ has $(d + 1)$ real roots.$\implies a_1,a_2,...,a_{d+1}$ if $d \geq 5$ $1)k \leq d-3$ $\implies \sum a_i = 0,\sum a_i \cdot a_j =0 \implies \sum(a_i)^2 =0 \implies$ Contradiction. $$a_1 \cdot a_2 \cdot a_3 .... \cdot a_{d+1} = -1$$$2)$ by $d-2 \geq 3$ $\implies \sum \frac{1}{a_i} = 0, \sum \frac{1}{a_i \cdot (a_j)} = 0$ $\implies \sum \frac{1}{(a_i)^2} = 0 \implies$ for $k=4$ $x^4+x^3-4x^2+x+1=(x-1)^2\cdot (x^2+3x+1)$ Answer:$\boxed {k=4}$