To prove that for all $n \ge 4$ we have $8n^4+23 < 7^n$.
We use induction to prove the result.
Base case: $n = 4$ the result is true.
Now we assume for $n = k$ i.e. we have $8k^4+23 < 7^k$.
Next, we prove for $n = k+1$, i.e. we have to show that $8(k+1)^4 + 23 < 7^{k+1}$.
Expanding, $8k^4 + 32k^3 +48k^2+32k+8 +23< 7 \times 7^k$
Thus it is sufficient to prove that $32k^3 +48k^2+32k+8 < 6 \times 7^k$
or, $4 [(2k+1)^3 + (2k+1)] < 6 \times 7^k$
After this do we need to apply induction again or is there any easy method?
Note that since $k\geq4$, we have $$ 32k^3\leq 8k^4\\ 48k^2\leq 3k^4\\ 32k\leq \frac12k^4< k^4 $$ Using this along with the induction hypothesis again, we get $$ 32k^3+48k^2+32k+8< 12k^4+8\\ < 6(8k^4+23)\leq 6\times 7^k $$