Why does $(2n+1)^{d}-(2n-1)^{d}\leq d(2n+1)^{d-1}$ hold? Why does the inequality $(2n+1)^{d}-(2n-1)^{d}\leq d(2n+1)^{d-1}$ hold for $d,\;n \in \mathbb N$? I am sure I could do the above via induction. Nonetheless, I would like to know whether there are other ways to go about this without using induction?

Something's wrong, because plugging in $n=1$, $d=2$ I get
$$3^2 - 1^2 \le 2 \cdot 3^1
$$
which says $8 \le 6$. Oops.
On the other hand, with another factor of $2$ on the right hand side it seems to be true. Substituting $a=2n+1$ and $b=2n-1$ into the formula
$$a^d-b^d=(a-b)(a^{d-1} + a^{d-2}b + ... + a b^{d-2} + b^{d-1})
$$
I get
\begin{align*}
(2n+1)^d - (2n-1)^d &= 2 \cdot \bigl((2n+1)^{d-1} + (2n+1)^{d-2} (2n-1) + ... \\
 & \qquad\qquad+ (2n+1)(2n-1)^{d-2} + (2n-1)^{d-1}\bigr) \\
 &\le 2d (2n+1)^{d-1}
\end{align*}
