Induction step: $5 + 5n \leq {n}^2$ for $n \geq 6$ Prove by mathematical induction that $5 + 5n \leq {n}^2 $ for all integers $n\geq 6$.
Step 1: Base case
Suppose $n = 6$, hence $5 + 5(6) \leq {6}^2 = 35 \leq 36$
We proved that base case is true as 35 is less than or equal to 36.
Step 2: Induction step
We claim that k is true for some integer more than or equal to 6, therefore $5 + 5k \leq {k}^2$ (*)
We now need to prove that k+1 claim is true and that is $5 + 5(k + 1) \leq {(k + 1)}^2$
I am stuck at this step. Somehow I am unable to sub in my claim k which is the (*) equation into my k+1 equation correctly. Is the step up to this point correct?
I have tried expanding out the k+1 equation for the LHS to get $5k + 10$ but it looks absolutely wrong. If I do it to the RHS, I get $7k + 6$ which looks wrong too although I am able to sub in ${k}^2$ to the RHS equation.
Can someone please tell me how to proceed from here on? 

For the induction step, you assume $5+5k\le k^2$. 
Now add $2k+1$ to both sides:
$$5+5k+(2k+1)\le (k+1)^2.$$
But $k\ge 6$, so certainly $2k+1\ge 5$.
Hence $5+5(k+1)\le (k+1)^2$.
