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!

Let $f(n)=\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}$
$$f(n+1)-f(n)=\frac{(n+1)^5}5+\frac{(n+1)^3}3+\frac{7(n+1)}{15}-\left(\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}\right)$$
$$=\frac{(n+1)^5-n^5}5+\frac{(n+1)^3-n^3}3+\frac{7(n+1)-7n}{15}$$
$$=\frac{5n^4+10n^3+10n^2+5n+1}5+\frac{3n^2+3n+1}3+\frac{7}{15}$$
$$=(n^4+2n^3+2n^2+n)+(n^2+n)+\left(\frac15+\frac13+\frac7{15}\right)$$ (using Binomial Theorem)
Now, as you have shown $\frac15+\frac13+\frac7{15}=1$
$$\text{So, if }f(n)=\frac{n^5}5+\frac{n^3}3+\frac{7n}{15} \text{ is an integer for integer } n$$
$$\text{ so will be }f(n+1)=\frac{(n+1)^5}5+\frac{(n+1)^3}3+\frac{7(n+1)}{15}$$
This problem can be extended to more primes and proved using the fact that prime $p$ divides $\binom pr$ for $1\le r\le p-1$
I can not keep back a non-inductive proof:
$$\frac{n^5}5+\frac{n^3}3+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+\frac{7n}{15}+\frac n3+\frac n5=\frac{n^5-n}5+\frac{n^3-n}3+n$$
Now, using Fermat's Little Theorem, $n^p-n\equiv0\pmod p$ for any integer $n$ and prime $p$
Similarly, $$\frac{n^7-n}7+\frac{n^5-n}5+\frac{n^3-n}3+n-\left(\frac n7+\frac n3+\frac n5\right)=\frac{n^7}7+\frac{n^5}5+\frac{n^3}3+\frac{34n}{105}$$ will be an integer
