Proof by induction that the sum of terms is integer I'm having some trouble in order to solve this induction proof.

Proof that $\forall{n} \in \mathbb{N}$ the number $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n$ is an integer. 

I've tried proving this by induction, but I've not succeed so far. What I did was:

Let's assume that $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$. So, by induction,  
$n=1$
  $\frac{1}{5} + \frac{1}{3} + \frac{7}{15} = 1$ and $1 \in \mathbb{Z}$.
Inductive step
  I want to prove that if $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$ then $\frac{1}{5}(n+1)^5+\frac{1}{3}(n+1)^3 + \frac{7}{15}(n+1) = l$, with $l \in \mathbb{Z}$

And then I'm stuck. I've tried developing every binomial but, for example, for $\frac{1}{5}(n+1)^5$ I get a $\frac{1}{5}n^4$ that is not in the inductive hypothesis, so therefore, I cannot get rid of it and it's not an integer, so I cannot conclude that $l \in \mathbb{Z}$.
Any help or ideas from where I can follow? Thanks in advance!

Hint
$$5|(n+1)^5-n^5-1 \,.$$
$$3|(n+1)^3-n^3-1 \,.$$
Write $a=\frac{(n+1)^5-n^5-1}{5} \,;\, b=\frac{(n+1)^3-n^3-1}{3}$, and express $P(n+1)-P(n)$ in terms of $a,b$, where $P(n)=\frac{1}{5}n^5+\frac{1}{3}n³ + \frac{7}{15}n$.
To prove the Hint, you can use either Fermat Little Theorem or binomial Theorem.
