While there have been many good proofs suggested in answers or comments, that are definitely the right way to do this for more general subgroups, this particular example is sufficiently straightforward that an elementary argument seems easiest. Put $c_i=b^{-i}ab^i$ for $i\leq 0\leq d-1$. Then $c_0,\ldots, c_{d-1}$ generate your subgroup. I claim by induction on length that if $w$ is a nontrivial reduced word in the free group on $d$ generators $x_0,\ldots, x_{d-1}$, then evaluating that word on $c_0,\ldots, c_{d-1}$ is nontrivial. More precisely, I claim that if the last letter of $w$ is $x_i^{\epsilon}$ with $\epsilon=\pm 1$, then $w(c_0,\ldots, c_{d-1})$ ends in $b^{i}$.

This is clear if $w$ is a single letter. Otherwise, by induction $w=ux_i^{\epsilon}$ where $u$ is reduced. Since the last letter of $u$ is not $x_i^{-\epsilon}$, we deduce by induction that $u(c_0,\ldots, c_{d-1})$ does not end in $b^{i}$. Therefore, we are unable to cancel the $a^{\pm 1}$ when we multiply $u(c_0,\ldots, c_{d-1})$ by $c_i^{\epsilon}$ (a $b^{\pm 1}$ from $c_i$ or the last letter plugged into $u$ survives), and so the reduced form of $u(c_0,\ldots, c_{d_1})c_i^\epsilon$ will end in $b^{i}$ as required.

While there have been many good proofs suggested in answers or comments, that are definitely the right way to do this for more general subgroups, this particular example is sufficiently straightforward that an elementary argument seems easiest. Put $c_i=b^{-i}ab^i$ for $i\leq 0\leq d-1$. Then $c_0,\ldots, c_{d-1}$ generate your subgroup. I claim by induction on length that if $w$ is a nontrivial reduced word in the free group on $d$ generators $x_0,\ldots, x_{d-1}$, then evaluating that word on $c_0,\ldots, c_{d-1}$ is nontrivial. More precisely, I claim that if the last letter of $w$ is $x_i^{\epsilon}$ with $\epsilon=\pm 1$, then $w(c_0,\ldots, c_{d-1})$ ends in $a^{\epsilon}b^{i}$.

This is clear if $w$ is a single letter. Otherwise, by induction $w=ux_i^{\epsilon}$ where $u$ is reduced. Since the last letter of $u$ is not $x_i^{-\epsilon}$, we deduce by induction that $u(c_0,\ldots, c_{d-1})$ ends in $a^{\pm 1}b^{j}$ with $j\neq i$ unless the last letter is $x_i^{\epsilon}$, in which case the reduced form ends in $a^{\epsilon}b^i$. Therefore, when we multiply $u(c_0,\ldots, c_{d-1})$ by $c_i^{\epsilon}$, we either get something ending in $b^{j-i}a^{\epsilon}b^i$ with $j\neq i$ or the reduced form of $u(c_0,\ldots, c_{d_1})c_i^\epsilon$ will end in $a^{\epsilon }a^{\epsilon}b^{i}$ as required.

While there have been many good proofs suggested in answers or comments, that are definitely the right way to do this for more general subgroups, this particular example is sufficiently straightforward that an elementary argument seems easiest. Put $c_i=b^{-i}ab^i$ for $i\leq 0\leq d-1$. Then $c_0,\ldots, c_{d-1}$ generate your subgroup. I claim by induction on length that if $w$ is a nontrivial reduced word in the free group on $d$ generators $x_0,\ldots, x_{d-1}$, then evaluating that word on $c_0,\ldots, c_{d-1}$ is nontrivial. More precisely, I claim that if the last letter of $w$ is $x_i^{\epsilon}$ with $\epsilon=\pm 1$, then $w(c_0,\ldots, c_{d-1})$ ends in $b^{i}$.

This is clear if $w$ is a single letter. Otherwise, by induction $w=ux_i^{\epsilon}$ where $u$ is reduced. Since the last letter of $u$ is not $x_i^{-\epsilon}$, we deduce by induction that $u(c_0,\ldots, c_{d-1})$ does not end in $b^{i}$. Therefore, we are unable to cancel the $a^{\pm 1}$ when we multiply $u(c_0,\ldots, c_{d-1})$ by $c_i^{\epsilon}$ (a $b^{\pm}$ from $c_i$ or the last letter plugged into $u$ survives), and so the reduced form of $u(c_0,\ldots, c_{d_1})c_i^\epsilon$ will end in $b^{i}$ as required.

While there have been many good proofs suggested in answers or comments, that are definitely the right way to do this for more general subgroups, this particular example is sufficiently straightforward that an elementary argument seems easiest. Put $c_i=b^{-i}ab^i$ for $i\leq 0\leq d-1$. Then $c_0,\ldots, c_{d-1}$ generate your subgroup. I claim by induction on length that if $w$ is a nontrivial reduced word in the free group on $d$ generators $x_0,\ldots, x_{d-1}$, then evaluating that word on $c_0,\ldots, c_{d-1}$ is nontrivial. More precisely, I claim that if the last letter of $w$ is $x_i^{\epsilon}$ with $\epsilon=\pm 1$, then $w(c_0,\ldots, c_{d-1})$ ends in $b^{i}$.

This is clear if $w$ is a single letter. Otherwise, by induction $w=ux_i^{\epsilon}$ where $u$ is reduced. Since the last letter of $u$ is not $x_i^{-\epsilon}$, we deduce by induction that $u(c_0,\ldots, c_{d-1})$ does not end in $b^{i}$. Therefore, we are unable to cancel the $a^{\pm 1}$ when we multiply $u(c_0,\ldots, c_{d-1})$ by $c_i^{\epsilon}$ (a $b^{\pm 1}$ from $c_i$ or the last letter plugged into $u$ survives), and so the reduced form of $u(c_0,\ldots, c_{d_1})c_i^\epsilon$ will end in $b^{i}$ as required.