We know
$$\lim_{n \to \infty} \frac{F_{n+1}}{F_n}=\Phi$$
by continuity of logarithm function, this implies
$$\lim_{n \to \infty} (log{F_{n+1}}-log{F_n})=log\Phi$$
By definition, $\forall{\epsilon \gt 0}, \exists{N\in {\mathbb N}}$ such that $n \ge N $ implies $|log{F_{n+1}}-log{F_n}-log\Phi|\lt \epsilon_1$
So by triangle inequality,
$$|(\sum_{k=N}^{n}\frac{1}{n-N+1}(logF_{k+1}-logF_k))-log\Phi|\le \frac{1}{n-N+1}\sum_{k=N}^{n}|logF_{k+1}-logF_k-log\Phi| \lt \epsilon_1
$$
Let $$\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-log\Phi|=A$$
Then $$|\sum_{k=1}^{n}(logF_{k+1}-logF_k-log\Phi)| \le \sum_{k=1}^{n} |logF_{k+1}-logF_k-log\Phi|\;,$$
which can be split into
$$\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-log\Phi|+\sum_{k=N}^{n} |logF_{k+1}-logF_k-log\Phi|$$
Hence, we have $$\frac{1}{n-N+1}(\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-\Phi|+\sum_{k=N}^{n} |logF_{k+1}-logF_k-log\Phi|)\lt \frac{A}{n-N+1}+\epsilon_1$$
Now, fix $N=N_0$. Since A is also fixed, $A \gt 0, \forall{\epsilon_2 \gt 0}, \exists{N_1 \in{\mathbb N}}$ such that $n \ge N_1 \;\text{implies} \; (n-N_0+1)\epsilon_2\lt A, \;\text{so}\; \dfrac{A}{n-N_0+1}\lt \epsilon_2.$
Hence, $\forall{n \ge N_1},$
$$|(\sum_{k=1}^{n}\frac{1}{n}(logF_{k+1}-logF_k))-log\Phi|\lt|\sum_{k=1}^{n}\frac{1}{n-N_0+1}(logF_{k+1}-logF_k-log\Phi)|\lt \epsilon_1+\epsilon_2$$
$$|\frac{1}{n}logF_{n+1}-log\Phi|=|(\sum_{k=1}^{n}\frac{1}{n}(logF_{k+1}-logF_k))-log\Phi|\lt \varepsilon_1$$
Therefore, by definition,
$$lim_{n \to \infty}\frac{1}{n}logF_{n+1}=lim_{n \to \infty}log\sqrt[n]{F_{n+1}}=log\Phi$$
By the continuity of the exponential function $e^x$, we get
$$lim_{n\to \infty} \sqrt[n]{F_{n+1}}=\Phi$$
Hence proven.
Remark: "Introduction to Calculus and Analysis I" by Richard Courant and Fritz John,page 114,SECTION 1.7, exercise $^*10$: "If $a_n \gt 0, \;\text{and} \; lim_{n\to \infty}\dfrac{a_{n+1}}{a_n}=L,\;\text{then}\;lim_{n\to \infty}\sqrt[n]{a_n}=L."$