Step 1: Look closely at the angle.
We want $\tan\left(\dfrac{7\pi}{8}\right)$. Notice that $\dfrac{7\pi}{8}$ is just a little less than $\pi$, so we can write it as a difference.
Step 2: Split the angle nicely.
Write $\dfrac{7\pi}{8}=\pi-\dfrac{\pi}{8}$. This lets us use a friendly reduction identity.
Step 3: Apply the reduction formula.
Since $\tan(\pi-\theta)=-\tan\theta$, we get \[ \tan\left(\frac{7\pi}{8}\right)=-\tan\left(\frac{\pi}{8}\right). \] Step 4: Recall the half-angle value.
Here $\dfrac{\pi}{8}=22.5^\circ$, which is half of $45^\circ$. The standard known value is $\tan 22.5^\circ=\sqrt{2}-1$.
Step 5: Quick check of that value.
Using $\tan\dfrac{\theta}{2}=\dfrac{1-\cos\theta}{\sin\theta}$ with $\theta=45^\circ$ gives $\dfrac{1-\frac{1}{\sqrt2}}{\frac{1}{\sqrt2}}=\sqrt2-1$, confirming $\tan\dfrac{\pi}{8}=\sqrt{2}-1$.
Step 6: Put it all together.
Therefore \[ \tan\left(\frac{7\pi}{8}\right)=-(\sqrt{2}-1)=1-\sqrt{2}. \] So the value is $1-\sqrt{2}$.
\[ \boxed{1-\sqrt{2}} \]