Prove $5^n +5 <5^{n+1}$ $∀n∈N$
Base Case: $n=1$
$\implies 5^1 +5 <25$
$\implies 10<25$ ; holds true
Induction hypothesis: Suppose $5^k +5 < 5^{k+1}$ is true for k∈N
Then;
$\implies 5^{k+1} +5 < 5^{k+2}$
$\implies 5\cdot 5^k +5 < 25*5^k$
I don't know how to proceed after this step.
$5^{k+1} + 5 < 5^{k+2} = 5(5^{k} + 1 < 5^{k+1})$. As $5^{k} + 1 < 5^{k} + 5 < 5^{k+1}$(from the inductive hypothesis), you have your answer.
I often find it helpful in a problem like your own to simply rewrite "the left hand side" so that it is immediately set up so you can apply the inductive hypothesis. Your remaining task, usually, is to then make sure you did not change the value.
In this particular problem, for example, you need to somehow obtain the inequality $5^{k+1}+5<5^{k+2}$ by using the inductive hypothesis (i.e. $5^k+5<5^{k+1}$). To do this, I would immediately write the following (after having shown the base case, of course): $$ 5^{k+1}+5\;=\;?\;(5^k+5)\;\pm\;?<\;?\;(5^{k+1})\;\pm\;?. $$ This is probably unclear at the moment, but actually filling in the details should clarify: $$ \begin{align*} 5^{k+1}+5&=5(5^k+5)-20\tag{set up to use ind. hyp.}\\[0.5em] &< 5(5^{k+1})-20\tag{by ind. hyp.}\\[0.5em] &<5^{k+2}.\tag{simplify} \end{align*} $$ Does that make sense? The goal with many of these problems is to use the inductive hypothesis effectively. You can make your work easier if you set it up at the outset so you can use the inductive hypothesis right away and then work towards your desired conclusion (as illustrated above).
You don't really need induction: $$ 5^n + 5 \le 5^n + 5^n = 2\cdot 5^n < 5\cdot 5^n = 5^{n+1} $$ but you can justify $5 \le 5^n$ by induction if you must.
$$5^{k+1}+5=5\times 5^k+5<5(5^{k+1}-5)+5=5^{k+2}-25+5<5^{k+2}$$ Notice that, by the induction step, we have $$5^k+5<5^{k+1}\implies 5^k<5^{k+1}-5$$
Well, we need to prove $5^n+5<5^{n+1}$
$5^n+5<5^{n+1}\implies 5<5^n\cdot 4\implies 1<5^{n-1}\cdot4$, which is obvious. Use induction or do whatever you want to do.
If you want to use induction, note that base step is true and if you assume that $5^{k-1}\cdot 4>1 $ then it is certainly true that $5^{k}\cdot 4=5\times (5^{k-1}\cdot 4)>1$