In Expected number of rolls until a number appears $k$ consecutive times, the formula was given to be $E[N] = \frac{6^k - 1}{5}$. You can prove this formula using induction like in the accepted answer.
I am wondering how do you actually derive this expression from scratch?
Denote by $e$ the expected value of getting, say, $6$ on a die roll, then $e =\frac{1}{p} = 6\;$ by the geometric distribution.
The final result is just the sum of the G.P. representing repeating the result of the first toss $(k-1)$ more times in succession
$1 + e + e^2 + ....e^{k-1} = \frac {e^k-1}{e-1} =\frac{6^k-1}{6-1}$
Added: Formula derivation
I am changing the symbol $e$ used above to $e_1$, to specify that it means the expected value for getting one six, $e_1 = 6$ from the geometric distribution
From there, we have $\frac16$ chance of getting double $6$, else back to scratch with a wasted throw, so $\;e_2 = e_1 +\frac16\cdot1 + \frac56(e_2 +1) \to\; e_2 = 6(e_1+1)$
We compute for $(k-1)$ more successive sixes using recursion
$\displaylines {e_2 = 6(e_1+1)=6+6^2\\e_3= 6(e_2+1) = 6 + 6^2+6^3\\e_4 = 6(e_3+1) = 6 + 6^2 +6^3+6^4\\ ... ...\\e_k = 1+ e_{k-1} = 1+6+6^2+6^3 +...6^{k-1} = \frac{6^k-1}{6-1}}$