Let $L(n)$ be the number of strings with an alphabet of $K$ characters that don't end in $0$ and don't have two successive $0$'s. Let $M(n)$ be the number of strings with an alphabet of $K$ characters that end in $0$ and don't have two successive $0$'s. Then $L(1)=K-1, M(1)=0$(because of no leading zero)$, L(n)=(K-1)(L(n-1)+M(n-1)), M(n)=L(n-1)$ the third because you can add any non-zero character to an acceptable string to get an acceptable string that doesn't end in zero, and you can add a zero to any acceptable string that doesn't end in zero to get an acceptable string that ends in zero. This makes a $2 \times 2$ matrix that takes $(L(n)\ M(n))^T$ to $(L(n+1)\ M(n+1))^T$ so you might find the eigenvalues and eigenvectors of that matrix. To get the final answer, you take $L(n)+M(n)$.
Added: to get a closed form we represent the recursion as $\begin {pmatrix} L(n) \\ M(n) \end {pmatrix}=\begin {pmatrix} 9 & 9 \\ 1 & 0 \end {pmatrix}\begin {pmatrix} L(n-1) \\ M(n-1) \end {pmatrix}$
This has eigenvalues $\lambda_{\pm}=\frac 32 (3 \pm \sqrt {13})$ with corresponding eigenvectors $\begin {pmatrix} 9+3\sqrt {13} \\ 1 \end {pmatrix},\begin {pmatrix} 9-3\sqrt {13} \\ 1 \end {pmatrix}$. Starting from $\begin {pmatrix} L(0) \\ M(0) \end {pmatrix}=\begin {pmatrix} K-1 \\ 0 \end {pmatrix}$ we get $\begin {pmatrix} L(n) \\ M(n) \end {pmatrix}= \frac{K-1}{6\sqrt{13}}\begin {pmatrix} 9+3\sqrt {13} \\ 1 \end {pmatrix}\lambda_+^n-\frac{K-1}{6\sqrt{13}}\begin {pmatrix} 9-3\sqrt {13} \\ 1 \end {pmatrix}\lambda_-^n$