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!

You want to try to write this expression in terms of the previous one.  So, start by expanding out the terms:
$$
\begin{align*}
\frac{1}{5}(n+1)^5&=\frac{1}{5}\sum_{k=0}^{5}\binom{5}{k}n^k\\
&=\frac{1}{5}(n^5+5n^4+10n^3+10n^2+5n+1)\\
&=\frac{1}{5}n^5+n^4+2n^3+2n^2+n+\frac{1}{5}
\end{align*}
$$
Do the same thing for $(n+1)^3$ and $(n+1)$.  Now, you can pick out the terms corresponding to the inductive hypothesis; all that's left is to show that the other remaining terms ALSO give you an integer.
