Assume on the contrary that $f_n$ is reducible, then write $$1+(x-1)(x-2) \ldots (x-n)=f_n(x)=g(x)h(x)$$ where $1 \leq \deg(g), \deg(h) \leq n-1$. Then $g(i)h(i)=1$ for $1 \leq i \leq n$, so $g(i)=h(i)= \pm 1$. This implies that $g(x)-h(x)$ is a polynomial of degree at most $n-1$, with $1, 2, \ldots , n$ as roots. Thus $g(x)-h(x)$ is identically $0$, so $f_n(x)=g(x)h(x)=g(x)^2$.
In my answer here, I show (after the first part about $P(n)$ always perfect square implying that $P(x)$ is a square of a polynomial) that $m+x(x+1) \ldots (x+k-1)$ is not a square of a polynomial, unless $k=4$ and $m=1$. This essentially finishes off the problem, since here we have $m=1, k=n$ and we may take $x \to x-n$.
Andreas Caranti's answer there also links to a proof of irreducibility of $f_n$ for $n>4$; I take the liberty of quoting him:
Some $10$ years ago a question was asked in a newsgroup (remember?), whether the polynomial
$$
x (x-1) (x-2) \dots (x - (k-1)) + 1
$$
is irreducible in $\mathbf{Z}[x]$ for $n > 4$. Small cases are easily dealt with, in particular for $n = 4$ you get
$$
x (x-1) (x-2) (x-3) + 1 = x^4 - 6 x^3 + 11 x^2 - 6 x + 1 = (x^2 - 3x +1)^2.
$$
Note that if you change $x$ to $-x$, you find a formula for your statement that $a(a+1)(a+2)(a+3) + 1$ is always the square of an integer.
I gave a proof of irreducibility, which appealed to Pythagorean triples in the end. It is written up in my Algebra notes, on page 76. These notes are in Italian, if needed I can easily translate them.
