Hint $\rm\ 5\! +\! 5^n < 5^{n+1}\!\!\iff\! 5 < 5^{n+1}\!-\!5^n = 4\cdot 5^n =: f(n).\,$ $\rm\:f(n)\:$ is increasing $\rm\: f(n\!+\!1)\ge f(n)\:$ hence a simple induction shows $\rm\:f(n) \ge f(1) > 5.$
Remark $\ $ This is a prototypical example of inequality telescoping. The hinted inductive proof - that an increasing function stays greater than its initial value - can be vividly viewed as collapsing $\rm\:f(n)\ge f(n\!-\!1)\ge\,\cdots\,f(2)\ge f(1)\:$ down to $\rm\:f(n)\ge f(1),\:$ using the transitivity of $\,\ge,\,$ just like one collapses the sections of a telescope down to one section.
Note that proving said lemma on increasing functions is actually easier then proving the simple case in your problem, because the special-case details no longer obscure the innate telescopic structure. Further, this lemma can be reused for many induction problems of similar type. Ditto for the higher-level conceptual understanding of how these types of induction work.
The above method of transforming an inductive proof into a much simpler telescopic proof is quite powerful, and very widely applicable, not only for inequalities but also for sums, products, etc. For many further examples see my prior posts on telescopy.